146.
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.
147.
main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are 'small-endian' machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.
148.
main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.
149.
#include ‹stdio.h›
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr < least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After 'ptr' reaches the end of the string the value pointed by 'str' is '\0'. So the value of 'str' is less than that of 'least'. So the value of 'least' finally is 0.
150.
main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of 'student' the member of type struct date is given. The compiler doesn't have the definition of date structure (forward reference is not allowed in C in this case) so it issues an error.
1-3
===
C Sample Question
Note : All the C sample programs are tested under Turbo C/C++ compilers.
It is assumed that,
* Programs run under DOS environment,
* The underlying machine is an x86 system,
* Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
Following are some C sample questions.
Predict the output or error(s) for the following:
1.
void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
2.
main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
3.
main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ).
4-6
===
C Sample Question
4.
main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.
5.
main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
6.
main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .
7.
main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression 'i++ && j++ && k++' is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for '0 || 0' combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
====
C Sample Question
8.
main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
9.
main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
10.
main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
11.
main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
===
C Sample Question
12.
main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
13.
#define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14.
main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
15.
#include‹stdio.h›
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 - 32), we get 77("M");
16. So we get the output 77 :: "M" (Ascii is 77).
#include‹stdio.h›
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
===
C Sample Question
17.
#include‹stdio.h›
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18.
#include‹stdio.h›
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure thecompiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.
19.
main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20.
main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.
21.
#define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
===
C Sample Question
22.
main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
* *p that is value at the location currently pointed by p will be taken
* ++*p the retrieved value will be incremented
* when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is 'h', which is changed to 'i' by executing ++*p and pointer moves to point, 'a' which is similarly changed to 'b' and so on. Similarly blank space is converted to '!'. Thus, we obtain value in p becomes “ibj!gsjfoet� and since p reaches '\0' and p1 points to p thus p1doesnot print anything.
23.
#include‹stdio.h›
#define a 10 main() { #define a 50 printf("%d",a); }
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.
24.
#define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem
25.
main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
===
C Sample Question
26.
main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
27.
enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation: enum assigns numbers starting from 0, if not explicitly defined.
28.
void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
29.
main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.
30.
main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.
===
C Sample Question
31.
main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.
32.
main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
33.
void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
34.
void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
35.
#include‹stdio.h›
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
==
C Sample Question
36.
main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.
37.
#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
38.
main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).
39.
#include‹stdio.h›
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32". i.e. (11+98-32)=77("M");
40.
#include‹stdio.h›
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration
==
C Sample Question
41.
#include‹stdio.h›
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
42.
main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.
43.
main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.
44.
main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.
45.
main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
==
C Sample Question
46.
main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf("%u %u %u %d \n",a,*a,**a,***a);
printf("%u %u %u %d \n",a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.
47.
main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf("%d" ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf("%d " ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
48.
main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
*ptr++;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
*++ptr;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
++*ptr;
printf("\n %d %d %d", ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr - p is value in ptr - starting location of array p, (1002 - 1000) / (scaling factor) = 1, *ptr - a = value at address pointed by ptr - starting value of array a, 1002 has a value 102 so the value is (102 - 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr - p = 2, *ptr - a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr - p = 3, *ptr - a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr - p = 1006 - 1000 = 3, *ptr - a = 108 - 100 = 4, **ptr = 4.
==
C Sample Question
49.
main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf("%s" ,(q+j));
for (j=0; j<3; j++) printf("%c" ,*(q+j));
for (j=0; j<3; j++) printf("%s" ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
M O U S E \0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
M T R A C K \0
The third input starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
50.
main( )
{
void *vp;
char ch = 'g', *cp = "goofy";
int j = 20;
vp = &ch;
printf("%c", *(char *)vp);
vp = &j;
printf("%d",*(int *)vp);
vp = cp;
printf("%s",(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is 'g'. Similarly the output from second printf is '20'. The third printf statement type casts it to print the string from the 4th value hence the output is 'fy'.
51.
main ( )
{
static char *s[ ] = {"black", "white", "yellow", "violet"};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf("%s",*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1-1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is 'ck'.
52.
main()
{
int i, n;
char *x = "girl";
n = strlen(x);
*x = x[n];
for(i=0; i‹n; ++i)
{
printf("%s\n",x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value "girl". The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location ('\0') to the first location. Now the string becomes "\0irl" . Now the printf statementprints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = '\0' hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e "irl" and the third time it prints "rl" and the last time it prints "l" and the loop terminates.
==
C Sample Question
53.
int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ‹file name›,‹line number›
Explanation: asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same.
After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of.
54.
main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).
55. What are the files which are automatically opened when a C file is executed?
Answer: stdin, stdout, stderr (standard input,standard output,standard error).
56. what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position of the file.
57.
main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.
58. What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF);
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.
59.
main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.
===
C Sample Question
60.
main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.
61.
main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.
62.
main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.
63.
#define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.
64.
main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
===
C Sample Question
65.
main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
66.
#define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.
67.
int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.
68.
main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.
69.
main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.
====
C Sample Question
70.
#include‹stdio.h›
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
71.
#include‹stdio.h›
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.
72.
#include‹stdio.h›
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.
73.
main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
74.
struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
===
C Sample Question
75.
struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
76.
main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
77.
main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.
78.
main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
79.
main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.
===
C Sample Question
80.
# include ‹stdio.h›
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
81.
# include‹stdio.h›
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
82.
#include‹stdio.h›
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.
83.
main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.
==
C Sample Question
84.
main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation: < blockquote>Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 sothe function returns value 1000.
85.
int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution will be,
t i x
4 0 -4
3 1 -2
2 2 0
86.
main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.
87.
main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.
===
C Sample Question
88. In the following pgm add a stmt in the function fun such that the address of 'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
89. What are the following notations of defining functions known as?
1. int abc(int a,float b)
{
/* some code */
}
2. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation (K&R Notation)
90.
main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.
91.
main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.
====
C Sample Question
>
92.
func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
93. The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed bythe function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. thereforethe function returns 0 which in turn is returned by the function 'process'.
void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.
94.
void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned inmain() it is printed after and preincrement.
95.
void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.
===
C Sample Question
96.
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
97.
void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.
98.
void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of time.
99.
void main()
{
int i=i++,j=j++,k=k++;
printf("%d%d%d",i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration. So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).
100.
void main()
{
static int i=i++, j=j++, k=k++;
printf("i = %d j = %d k = %d", i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
===
C Sample Question
101.
void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.
102.
main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534...
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.
103.
#include‹conio.h›
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule:
Check all control paths to write bug free code.
104.
main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
105.
#define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10
====
C Sample Question
106.
main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.
107.
main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.
108.
main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
109.
main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.
110.
void pascal f(int i,int j,int k)
{
printf(“%d %d %d�,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d�,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.
==
C Sample Question
111. What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
-128
Explanation: Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.
112.
main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
infinite loop
Explanation:
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of thefor loop.
113.
main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.
114. Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer:
Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
115. What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer:
Compiler error: Multiple declaration for error
Explanation:
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguishthe meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case? When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note:
the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer's convenience.
==
C Sample Question
116.
typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer:
1
Explanation:
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces). Typedef structerror{int warning, error, exception;}error; This error can be used only by preceding the error by struct kayword as in: struct error someError; typedef struct error{int warning, error, exception;}error; This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in : g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in: error g1; Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real programming don't use such overloading of names. It reduces the readability of the code. Possible doesn't mean that we should use it!
117.
#ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration int some = 0; effectively removed from the source code.
118.
#if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
0 0
Explanation:
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.
119.
void main()
{
if(~0 == (unsigned int)-1)
printf("You can answer this if you know how values are represented in memory");
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. -1 is represented in unsigned value as all 1's and so both are equal.
120.
int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer:
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand this.
====
C Sample Question
121.
main()
{
char *p = "ayqm";
printf("%c",++*(p++));
}
Answer:
b
122.
main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.
123.
main()
{
char *p = "ayqm";
char c;
c = ++*p++;
printf("%c",c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.
124.
int aaa() {printf("Hi");}
int bbb(){printf("hello");}
iny ccc(){printf("bye");}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing "bye".
125.
main()
{
int i=5;
printf("%d",i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.
====
C Sample Question
126.
main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = 'c' the string becomes, "%c\n". Since this string becomes the format string for printf and ASCII value of 65 is 'A', the same gets printed.
127.
void ( * abc( int, void ( *def) () ) ) ();
Answer:
abc is a ptr to a function which takes 2 parameters (a) an integer variable (b)a ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.
128.
main()
{
while (strcmp("some","some\0"))
printf("Strings are not equal\n");
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So "some" and "some\0" are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.
129.
main()
{
char str1[] = {'s','o','m','e'};
char str2[] = {'s','o','m','e','\0'};
while (strcmp(str1,str2))
printf("Strings are not equal\n");
}
Answer:
"Strings are not equal"
"Strings are not equal" ...
Explanation:
If a string constant is initialized explicitly with characters, '\0' is not appended automatically to the string. Since str1 doesn't have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a '\0'. So str1 and str2 are not the same, hence the result.
130.
main()
{
int i = 3;
for (;i++=0;) printf("%d",i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.
===
C Sample Question
131.
void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf("%d",*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf("%d",*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.
132.
void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf("%d", i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.
133.
main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
134.
1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
3. Same as 1.
135.
main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.
==
C Sample Question
136.
main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression's truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same. Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression's truth value becomes false and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
137.
main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.
138.
main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer: Compiler Error: switch expression not integral
Explanation: Switch statements can be applied only to integral types.
139.
main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.
140.
main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.
==
C Sample Question
141.
main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is thecharacter array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.
142.
#define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
143.
int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.
144.
main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].
145.
main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn't issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows, void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.
==
C Sample Question
146.
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.
147.
main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are 'small-endian' machines. Small-endian means that the lower order bytes are stored in the highermemory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.
148.
main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.
149.
#include ‹stdio.h›
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr < least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After 'ptr' reaches the end of the string the value pointed by 'str' is '\0'. So the value of 'str' is less than that of 'least'. So the value of 'least' finally is 0.
150.
main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of 'student' the member of type struct date is given. The compiler doesn't have the definition of date structure (forward reference is not allowed in C in this case) so it issues an error.
==
c++
===
C++ and OOPS Sample Question
Note : All the C sample programs are tested under Turbo C/C++ compilers.
It is assumed that,
* Programs run under DOS environment,
* The underlying machine is an x86 system,
* Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
Following are some C sample questions.
Predict the output or error(s) for the following:
1.
class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout « "The value is " « *ptr;
}
};
void SomeFunc(Sample x)
{
cout « "Say i am in someFunc " « endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in someFunc
Null pointer assignment(Run-time error)
Explanation:
As the object is passed by value to SomeFunc the destructor of the object is called when the control returns from the function. So when PrintVal is called it meets up with ptr that has been freed.The solution is to pass the Sample object by reference to SomeFunc:
void SomeFunc(Sample &x)
{
cout « "Say i am in someFunc " « endl;
}
because when we pass objects by refernece that object is not destroyed. while returning from the function.
2. Which is the parameter that is added to every non-static member function when it is called?
Answer: 'this' pointer
3.
class base
{
public:
int bval;
base(){ bval=0;}
};
class deri:public base
{
public:
int dval;
deri(){ dval=1;}
};
void SomeFunc(base *arr,int size)
{
for(int i=0; i‹size; i++,arr++)
cout«arr-›bval;
cout«endl;
}
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
}
Answer:
00000
01010
Explanation:
The function SomeFunc expects two arguments.The first one is a pointer to an array of base class objects and the second one is the sizeof the array.The first call of someFunc calls it with an array of bae objects, so it works correctly and prints the bval of all the objects. When Somefunc is called the second time the argument passed is the pointeer to an array of derived class objects and not the array of base class objects. But that is whatthe function expects to be sent. So the derived class pointer is promoted to base class pointer and the address is sent to the function . SomeFunc() knows nothing about this and just treats the pointer as an array of base class objects. So when arr++ is met, the size of base class object is taken into consideration and is incremented by sizeof(int) bytes for bval (the deri class objects have bval and dval as members and so is of size ›= sizeof(int)+sizeof(int) ).
====
C++ and OOPS Sample Question
4.
class base
{
public:
void baseFun(){ cout«"from base"«endl;}
};
class deri:public base
{
public:
void baseFun(){ cout« "from derived"«endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from base
Explanation:
As we have seen in the previous case, SomeFunc expects a pointer to a base class. Since a pointer to a derived class object is passed, it treats the argument only as a base class pointer and the corresponding base function is called.
5.
class base
{
public:
virtual void baseFun(){ cout«"from base"«endl;}
};
class deri:public base
{
public:
void baseFun(){ cout« "from derived"«endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from derived
Explanation:
Remember that baseFunc is a virtual function. That means that it supports run-time polymorphism. So the function corresponding to the derived class object is called.
6.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout «"a="«a «"*pa="«*pa «"ra"«ra ;
}
Answer:
Compiler Error: 'ra',reference must be initialized
Explanation:
Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned, whereas references can only be initialized. So this code issues an error.
7.
const int size = 5;
void print(int *ptr)
{
cout«ptr[0];
}
void print(int ptr[size])
{
cout«ptr[0];
}
void main()
{
int a[size] = {1,2,3,4,5};
int *b = new int(size);
print(a);
print(b);
}
Answer:
Compiler Error : function 'void print(int *)' already has a body
Explanation:
Arrays cannot be passed to functions, only pointers (for arrays, base addresses) can be passed. So the arguments int *ptr and int prt[size] have no difference as function arguments. In other words, both the functoins have the same signature and so cannot be overloaded. ===
==
C++ and OOPS Sample Question
8.
class some{
public:
~some()
{
cout«"some's destructor"«endl;
}
};
void main()
{
some s;
s.~some();
}
Answer :
some's destructor
some's destructor
Explanation:
Destructors can be called explicitly. Here 's.~some()' explicitly calls the destructor of 's'. When main() returns, destructor of s is called again, hence the result.
9.
#include ‹iostream.h›
class fig2d
{
int dim1;
int dim2;
public:
fig2d() { dim1=5; dim2=6;}
virtual void operator«(ostream & rhs);
};
void fig2d::operator«(ostream &rhs)
{
rhs «this->dim1«" "«this->dim2«" ";
}
/*class fig3d : public fig2d
{
int dim3;
public:
fig3d() { dim3=7;}
virtual void operator«(ostream &rhs);
};
void fig3d::operator«(ostream &rhs)
{
fig2d::operator «(rhs);
rhs«this -> dim3;
}
*/
void main()
{
fig2d obj1;
// fig3d obj2;
obj1 « cout;
// obj2 « cout;
}
Answer :
5 6
Explanation:
In this program, the « operator is overloaded with ostream as argument. This enables the 'cout' to be present at the right-hand-side. Normally, 'cout' is implemented as global function, but it doesn't mean that 'cout' is not possible to be overloaded as member function.
Overloading « as virtual member function becomes handy when the class in which it is overloaded is inherited, and this becomes available to be overrided. This is as opposed to global friend functions, where friend's are not inherited.
10.
class opOverload{
public:
bool operator==(opOverload temp);
};
bool opOverload::operator==(opOverload temp){
if(*this == temp ){
cout«"The both are same objects\n";
return true;
}
else{
cout«"The both are different\n";
return false;
}
}
void main(){
opOverload a1, a2;
a1= =a2;
}
Answer :
Runtime Error: Stack Overflow
Explanation:
Just like normal functions, operator functions can be called recursively. This program just illustrates that point, by calling the operator == function recursively, leading to an infinite loop.
===
C++ and OOPS Sample Question
11.
class complex{
double re;
double im;
public:
complex() : re(1),im(0.5) {}
bool operator==(complex &rhs);
operator int(){}
};
bool complex::operator == (complex &rhs){
if((this->re == rhs.re) && (this->im == rhs.im))
return true;
else
return false;
}
int main(){
complex c1;
cout« c1;
}
Answer :
Garbage value
Explanation :
The programmer wishes to print the complex object using output re-direction operator,which he has not defined for his lass.But the compiler instead of giving an error sees the conversion function and converts the user defined object to standard object and prints some garbage value.
12.
class complex{
double re;
double im;
public:
complex() : re(0),im(0) {}
complex(double n) { re=n,im=n;};
complex(int m,int n) { re=m,im=n;}
void print() { cout«re; cout«im;}
};
void main(){
complex c3;
double i=5;
c3 = i;
c3.print();
}
Answer :
5,5
Explanation :
Though no operator= function taking complex, double is defined, the double on the rhs is converted into a temporary object using the single argument constructor taking double and assigned to the lvalue.
13.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout «"a="«a «"*pa="«*pa «"ra"«ra ;
}
Answer :
Compiler Error: 'ra',reference must be initialized
Explanation :
Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned, whereas references can only be initialized. So this code issues an error.
==
C++ and OOPS Sample Question
14. Determine the output of the 'C++' Codelet.
class base
{
public :
out()
{
cout«"base ";
}
};
class deri{
public : out()
{
cout«"deri ";
}
};
void main()
{ deri dp[3];
base *bp = (base*)dp;
for (int i=0; i<3;i++)
(bp++)->out();
}
15. Justify the use of virtual constructors and destructors in C++.
16. Each C++ object possesses the 4 member fns,(which can be declared by the programmer explicitly or by the implementation if they are not available). What are those 4 functions?
17. What is wrong with this class declaration?
class something
{
char *str;
public:
something(){
st = new char[10]; }
~something()
{
delete str;
}
};
18. Inheritance is also known as -------- relationship. Containership as ________ relationship.
19. When is it necessary to use member-wise initialization list (also known as header initialization list) in C++?
20. Which is the only operator in C++ which can be overloaded but NOT inherited.
21. Is there anything wrong with this C++ class declaration?
class temp
{
int value1;
mutable int value2;
public :
void fun(int val)
const{
((temp*) this)->value1 = 10;
value2 = 10;
}
};
====
C++ and OOPS Sample Question
22. What is a modifier?
Answer:
A modifier, also called a modifying function is a member function that changes the value of at least one data member. In other words, an operation that modifies the state of an object. Modifiers are also known as 'mutators'.
23. What is an accessor?
Answer:
An accessor is a class operation that does not modify the state of an object. The accessor functions need to be declared as const operations
24. Differentiate between a template class and class template.
Answer:
Template class:
A generic definition or a parameterized class not instantiated until the client provides the needed information. It's jargon for plain templates.
Class template:
A class template specifies how individual classes can be constructed much like the way a class specifies how individual objects can be constructed. It'sjargon for plain classes.
25. When does a name clash occur?
Answer:
A name clash occurs when a name is defined in more than one place. For example., two different class libraries could give two different classes the same name. If you try to use many class libraries at the same time, there is a fair chance that you will be unable to compile or link the program because of name clashes.
26. Define namespace.
Answer:
It is a feature in c++ to minimize name collisions in the global name space. This namespace keyword assigns a distinct name to a library that allows otherlibraries to use the same identifier names without creating any name collisions. Furthermore, the compiler uses the namespace signature for differentiating the definitions.
27. What is the use of 'using' declaration.
Answer:
A using declaration makes it possible to use a name from a namespace without the scope operator.
28. What is an Iterator class?
Answer:
A class that is used to traverse through the objects maintained by a container class. There are five categories of iterators:
1. input iterators,
2. output iterators,
3. forward iterators,
4. bidirectional iterators,
5. random access.
An iterator is an entity that gives access to the contents of a container object without violating encapsulation constraints. Access to the contents is granted on a one-at-a-time basis in order. The order can be storage order (as in lists and queues) or some arbitrary order (as in array indices) or according to some ordering relation (as in an ordered binary tree). The iterator is a construct, which provides an interface that, when called, yields either the next element in the container, or some value denoting the fact that there are no more elements to examine. Iterators hide the details of access to and update of the elements of a container class.
The simplest and safest iterators are those that permit read-only access to the contents of a container class. The following code fragment shows how an iterator might appear in code:
cont_iter:=new cont_iterator();
x:=cont_iter.next();
while x/=none do
...
s(x);
...
x:=cont_iter.next();
end;
In this example, cont_iter is the name of the iterator. It is created on the first line by instantiation of cont_iterator class, an iterator class defined to iterate over some container class, cont. Succesive elements fromthe container are carried to x. The loop terminates when x is bound to some empty value. (Here, none)In the middle of the loop, there is s(x) an operation on x, the current element fromthe container. The next element of the container is obtained at the bottom of the loop.
===
C++ and OOPS Sample Question
29. List out some of the OODBMS available.
Answer:
* GEMSTONE/OPAL of Gemstone systems.
* ONTOS of Ontos.
* Objectivity of Objectivity inc.
* Versant of Versant object technology.
* Object store of Object Design.
* ARDENT of ARDENT software.
* POET of POET software.
30. List out some of the object-oriented methodologies.
Answer:
* Object Oriented Development (OOD) (Booch 1991,1994).
* Object Oriented Analysis and Design (OOA/D) (Coad and Yourdon 1991).
* Object Modelling Techniques (OMT) (Rumbaugh 1991).
* Object Oriented Software Engineering (Objectory) (Jacobson 1992).
* Object Oriented Analysis (OOA) (Shlaer and Mellor 1992).
* The Fusion Method (Coleman 1991).
31. What is an incomplete type?
Answer:
Incomplete types refers to pointers in which there is non availability of the implementation of the referenced location or it points to some location whose value is not available for modification.
Example:
int *i=0x400 // i points to address 400 *i=0; //set the value of memory location pointed by i.
Incomplete types are otherwise called uninitialized pointers.
32. What is a dangling pointer?
Answer: A dangling pointer arises when you use the address of an object after its lifetime is over. This may occur in situations like returning addresses of the automatic variables from a function or using the address of the memory block after it is freed.
33. Differentiate between the message and method.
Answer:
Message Method
Objects communicate by sending messages Provides response to a message. to each other.
A message is sent to invoke a method. It is an implementation of an operation.
34. What is an adaptor class or Wrapper class?
Answer:
A class that has no functionality of its own. Its member functions hide the use of a third party software component or an object with the non-compatible interface or a non- object- oriented implementation.
35. What is a Null object?
Answer:
It is an object of some class whose purpose is to indicate that a real object of that class does not exist. One common use for a null object is a return value from a member function that is supposed to return an object with some specified properties but cannot find such an object.
===
C++ and OOPS Sample Question
36. What is class invariant?
Answer:
A class invariant is a condition that defines all valid states for an object. It is a logical condition to ensure the correct working of a class. Class invariants must hold when an object is created, and they must be preserved under all operations of the class. In particular all class invariants are both preconditions and post-conditions for all operations or member functions of the class.
37. What do you mean by Stack unwinding?
Answer:
It is a process during exception handling when the destructor is called for all local objects between the place where the exception was thrown and where it is caught.
38. Define precondition and post-condition to a member function.
Answer:
Precondition:
A precondition is a condition that must be true on entry to a member function. A class is used correctly if preconditions are never false. An operation is not responsible for doing anything sensible if its precondition fails tohold.
For example, the interface invariants of stack class say nothing about pushing yet another element on a stack that is already full. We say that isful() is a precondition of the push operation.
Post-condition:
A post-condition is a condition that must be true on exit from a member function if the precondition was valid on entry to that function. A class is implemented correctly if post-conditions are never false.
>For example, after pushing an element on the stack, we know that isempty() must necessarily hold. This is a post-condition of the push operation.
39. What are the conditions that have to be met for a condition to be an invariant of the class?
Answer:
The condition should hold at the end of every constructor.
The condition should hold at the end of every mutator(non-const) operation.
40. What are proxy objects?
Answer:
Objects that stand for other objects are called proxy objects or surrogates.
Example:
template
class Array2D
{
public:
class Array1D
{
public:
T& operator[] (int index);
const T& operator[] (int index) const;
...
};
Array1D operator[] (int index);
const Array1D operator[] (int index) const;
...
};
The following then becomes legal:
Array2Ddata(10,20);
........
cout«data[3][6]; // fine
Here data[3] yields an Array1D object and the operator [] invocation on that object yields the float in position(3,6) of the original two dimensional array. Clients of the Array2D class need not be aware of the presence of the Array1D class. Objects of this latter class stand for one-dimensional array objects that, conceptually, do not exist for clients of Array2D. Such clients program as if they were using real, live, two-dimensional arrays. Each Array1D object stands for a one-dimensional array that is absent from a conceptual model used by the clients of Array2D. In the above example, Array1D is a proxy class. Its instances stand for one-dimensional arrays that, conceptually, do not exist.
41. Name some pure object oriented languages.
Answer:
* Smalltalk,
* Java,
* Eiffel,
* Sather.
42. Name the operators that cannot be overloaded.
Answer:
sizeof . .* .-> :: ?:
==
C++ and OOPS Sample Question
43. What is a node class?
Answer:
A node class is a class that,
* relies on the base class for services and implementation,
* provides a wider interface to te users than its base class,
* relies primarily on virtual functions in its public interface
* depends on all its direct and indirect base class
* can be understood only in the context of the base class
* can be used as base for further derivation
* can be used to create objects.
* A node class is a class that has added new services or functionality beyond the services inherited from its base class.
44. What is an orthogonal base class?
Answer:
If two base classes have no overlapping methods or data they are said to be independent of, or orthogonal to each other. Orthogonal in the sense means that two classes operate in different dimensions and do not interfere with each other in any way. The same derived class may inherit such classes with no difficulty.
45. What is a container class? What are the types of container classes?
Answer:
A container class is a class that is used to hold objects in memory or external storage. A container class acts as a generic holder. A container class has a predefined behavior and a well-known interface. A container class is a supporting class whose purpose is to hide the topology used for maintaining the list of objects in memory. When a container class contains a group of mixed objects, the container is called a heterogeneous container; when the container is holding a group of objects that are all the same, the container is called a homogeneous container.
46. What is a protocol class?
Answer:
An abstract class is a protocol class if:
* it neither contains nor inherits from classes that contain member data, non-virtual functions, or private (or protected) members of any kind.
* it has a non-inline virtual destructor defined with an empty implementation,
* all member functions other than the destructor including inherited functions, are declared pure virtual functions and left undefined.
47. What is a mixin class?
Answer:
A class that provides some but not all of the implementation for a virtual base class is often called mixin. Derivation done just for the purpose of redefining the virtual functions in the base classes is often called mixin inheritance. Mixin classes typically don't share common bases.
48. What is a concrete class?
Answer:
A concrete class is used to define a useful object that can be instantiated as an automatic variable on the program stack. The implementation of a concrete class is defined. The concrete class is not intended to be a base class and no attempt to minimize dependency on other classes in theimplementation or behavior of the class.
49. What is the handle class?
Answer:
A handle is a class that maintains a pointer to an object that is programmatically accessible through the public interface of the handle class.
Explanation:
In case of abstract classes, unless one manipulates the objects of these classes through pointers and references, the benefits of the virtual functions are lost. User code may become dependent on details ofimplementation classes because an abstract type cannot be allocated statistically or on the stack without its size being known. Using pointers or references implies that the burden of memory management falls on the user. Another limitation ofabstract class object is of fixed size. Classes however are used to represent concepts that require varying amounts of storage to implement them.
A popular technique for dealing with these issues is to separate what is used as a single object in two parts: a handle providing the user interface and a representation holding all or most of the object's state. The connection between the handle and the representation is typically a pointer in the handle. Often, handles have a bit more data than the simple representation pointer, but not much more. Hence the layout of the handle is typically stable, even when the representation changes and also that handles are small enough to move around relatively freely so that the user needn't use the pointers and the references.
====
C++ and OOPS Sample Question
50. What is an action class?
Answer:
The simplest and most obvious way to specify an action in C++ is to write a function. However, if the action has to be delayed, has to be transmitted 'elsewhere' before being performed, requires its own data, has to be combined with other actions, etc then it often becomes attractive to provide the action in the form of a class that can execute the desired action and provide other services as well. Manipulators used with iostreams is an obvious example.
Explanation:
A common form of action class is a simple class containing just one virtual function.
class Action
{
public:
virtual int do_it( int )=0;
virtual ~Action( );
}
Given this, we can write code say a member that can store actions for later execution without using pointers to functions, without knowing anything about the objects involved, and without even knowing the name of the operation it invokes. For example: class write_file : public Action
{
File& f;
public:
int do_it(int)
{
return fwrite( ).suceed( );
}
};
class error_message: public Action
{
response_box db(message.cstr( ),"Continue","Cancel","Retry");
switch (db.getresponse( ))
{
case 0: return 0;
case 1: abort();
case 2: current_operation.redo( );return 1;
}
};
A user of the Action class will be completely isolated from any knowledge of derived classes such as write_file and error_message.
51. When can you tell that a memory leak will occur?
Answer:
A memory leak occurs when a program loses the ability to free a block of dynamically allocated memory.
52. What is a parameterized type?
Answer:
A template is a parameterized construct or type containing generic code that can use or manipulate any type. It is called parameterized because an actual type is a parameter of the code body. Polymorphism may be achieved through parameterized types. This type of polymorphism is called parameteric polymorphism. Parameteric polymorphism is the mechanism by which the same code is used on different types passed as parameters.
53. Differentiate between a deep copy and a shallow copy?
Answer:
Deep copy involves using the contents of one object to create another instance of the same class. In a deep copy, the two objects may contain ht same information but the target object will have its own buffers and resources. the destruction of either object will not affect the remaining object. The overloaded assignment operator would create a deep copy of objects. Shallow copy involves copying the contents of one object into another instance of the same class thus creating a mirror image. Owing to straight copying of references and pointers, the two objects will share the same externally contained contents of the other object to be unpredictable.
Explanation:
Using a copy constructor we simply copy the data values member by member. This method of copying is called shallow copy. If the object is a simple class, comprised of built in types and no pointers this would be acceptable. Thisfunction would use the values and the objects and its behavior would not be altered with a shallow copy, only the addresses of pointers that are members are copied and not the value the address is pointing to. The data values of the object would then be inadvertently altered by thefunction. When the function goes out of scope, the copy of the object with all its data is popped off the stack.
If the object has any pointers a deep copy needs to be executed. With the deep copy of an object, memory is allocated for the object in free store and the elements pointed to are copied. A deep copy is used for objects that are returned from afunction.
54. What is an opaque pointer?
Answer:
A pointer is said to be opaque if the definition of the type to which it points to is not included in the current translation unit. A translation unit is the result of merging an implementation file with all its headers and header files.
==
C++ and OOPS Sample Question
55. What is a smart pointer?
Answer:
A smart pointer is an object that acts, looks and feels like a normal pointer but offers more functionality. In C++, smart pointers are implemented as template classes that encapsulate a pointer and override standard pointer operators. They have a number of advantages over regular pointers. They are guaranteed to be initialized as either null pointers or pointers to a heap object. Indirection through a null pointer is checked. No delete is ever necessary. Objects are automatically freed when the last pointer to them has gone away. One significant problem with these smart pointers is that unlike regular pointers, they don't respect inheritance. Smart pointers are unattractive for polymorphic code. Given below is an example for the implementation of smart pointers.
Example:
template ‹class X›
class smart_pointer
{
public:
smart_pointer(); // makes a null pointer
smart_pointer(const X& x) // makes pointer to copy of x
X& operator *( );
const X& operator*( ) const;
X* operator->() const;
smart_pointer(const smart_pointer ‹X› &);
const smart_pointer ‹X› & operator =(const smart_pointer‹X›&);
~smart_pointer();
private:
//...
};
This class implement a smart pointer to an object of type X. The object itself is located on the heap. Here is how to use it:
smart_pointer ‹employee› p= employee("Harris",1333);
Like other overloaded operators, p will behave like a regular pointer,
cout«*p;
p -> raise_salary(0.5);
56. What is reflexive association?
Answer:
The 'is-a' is called a reflexive association because the reflexive association permits classes to bear the is-a association not only with their super-classes but also with themselves. It differs from a 'specializes-from' as 'specializes-from' is usually used to describe the association between a super-class and a sub-class. For example: Printer is-a printer.
57. What is slicing?
Answer:
Slicing means that the data added by a subclass are discarded when an object of the subclass is passed or returned by value or from a function expecting a base class object.
Explanation:
Consider the following class declaration:
class base
{
...
base& operator =(const base&);
base (const base&);
}
void fun( )
{
base e=m;
e=m;
}
As base copy functions don't know anything about the derived only the base part of the derived is copied. This is commonly referred to as slicing. One reason to pass objects of classes in a hierarchy is to avoid slicing. Other reasons are to preserve polymorphic behavior and to gain efficiency.
58. What is name mangling?
Answer:
Name mangling is the process through which your c++ compilers give each function in your program a unique name. In C++, all programs have at-least a few functions with the same name. Name mangling is a concession to the fact that linker always insists on all function names being unique.
Example:
In general, member names are made unique by concatenating the name of the member with that of the class e.g. given the declaration:
class Bar
{
public:
int ival;
...
};
ival becomes something like:
// a possible member name mangling
ival__3Bar
Consider this derivation:
class Foo : public Bar
{
public:
int ival;
...
}
The internal representation of a Foo object is the concatenation of its base and derived class members.
// Pseudo C++ code
// Internal representation of Foo
class Foo
{
public:
int ival__3Bar;
int ival__3Foo;
...
};
Unambiguous access of either ival members is achieved through name mangling. Member functions, because they can be overloaded, require an extensive mangling to provide each with a unique name. Here the compiler generates the same name for the two overloaded instances(Their argument lists make their instances unique).
==
C++ and OOPS Sample Question
59. What are proxy objects?
Answer:
Objects that points to other objects are called proxy objects or surrogates. Its an object that provides the same interface as its server object but does not have any functionality. During a method invocation, it routes data to the true server object and sends back the return value to the object.
60. Differentiate between declaration and definition in C++.
Answer:
A declaration introduces a name into the program; a definition provides a unique description of an entity (e.g. type, instance, and function). Declarations can be repeated in a given scope, it introduces a name in a given scope. There must be exactly one definition of every object, function or class used in a C++ program.
A declaration is a definition unless:
* it declares a function without specifying its body,
* it contains an extern specifier and no initializer or function body,
* it is the declaration of a static class data member without a class definition,
* it is a class name definition,
* it is a typedef declaration.
A definition is a declaration unless:
* it defines a static class data member,
* it defines a non-inline member function.
61. What is cloning?
Answer:
An object can carry out copying in two ways i.e. it can set itself to be a copy of another object, or it can return a copy of itself. The latter process is called cloning.
62. Describe the main characteristics of static functions.
Answer:
The main characteristics of static functions include,
* It is without the a this pointer,
* It can't directly access the non-static members of its class
* It can't be declared const, volatile or virtual.
* It doesn't need to be invoked through an object of its class, although for convenience, it may.
63. Will the inline function be compiled as the inline function always? Justify.
Answer:
An inline function is a request and not a command. Hence it won't be compiled as an inline function always.
Explanation:
Inline-expansion could fail if the inline function contains loops, the address of an inline function is used, or an inline function is called in a complex expression. The rules for inlining are compiler dependent.
64. Define a way other than using the keyword inline to make a function inline.
Answer:
The function must be defined inside the class.
65. How can a '::' operator be used as unary operator?
Answer:
The scope operator can be used to refer to members of the global namespace. Because the global namespace doesn’t have a name, the notation :: member-name refers to a member of the global namespace. This can be useful for referring to members of global namespace whose names have been hidden by names declared in nested local scope. Unless we specify to the compiler in which namespace to search for a declaration, the compiler simple searches the current scope, and any scopes in which the current scope is nested, to find the declaration for the name.
66. What is placement new?
Answer:
When you want to call a constructor directly, you use the placement new. Sometimes you have some raw memory that's already been allocated, and you need to construct an object in the memory you have. Operator new's special version placement new allows you to do it.
class Widget
{
public :
Widget(int widgetsize);
...
Widget* Construct_widget_int_buffer(void *buffer,int widgetsize)
{
return new(buffer) Widget(widgetsize);
}
};
This function returns a pointer to a Widget object that's constructed within the buffer passed to the function. Such a function might be useful for applications using shared memory or memory-mapped I/O, because objects in such applications must be placed at specific addresses or in memory allocated by special routines.
==
C++ and OOPS Sample Question
67. What do you mean by analysis and design?
Analysis:
Basically, it is the process of determining what needs to be done before how it should be done. In order to accomplish this, the developer refers the existing systems and documents. So, simply it is an art of discovery.
Design:
It is the process of adopting/choosing the one among the many, which best accomplishes the users needs. So, simply, it is compromising mechanism.
68. What are the steps involved in designing?
Before getting into the design the designer should go through the SRS prepared by the System Analyst.
The main tasks of design are Architectural Design and Detailed Design.
In Architectural Design we find what are the main modules in the problem domain. In Detailed Design we find what should be done within each module.
69. What are the main underlying concepts of object orientation?
Objects, messages, class, inheritance and polymorphism are the main concepts of object orientation.
70. What do u meant by "SBI" of an object?
SBI stands for State, Behavior and Identity. Since every object has the above three.
* State:
It is just a value to the attribute of an object at a particular time.
* Behaviour:
It describes the actions and their reactions of that object.
* Identity:
An object has an identity that characterizes its own existence. The identity makes it possible to distinguish any object in an unambiguous way, and independently from its state.
71. Differentiate persistent & non-persistent objects?
Persistent refers to an object's ability to transcend time or space. A persistent object stores/saves its state in a permanent storage system with out losing the information represented by the object.
A non-persistent object is said to be transient or ephemeral. By default objects are considered as non-persistent.
72. What do you meant by active and passive objects?
Active objects are one which instigate an interaction which owns a thread and they are responsible for handling control to other objects. In simple words it can be referred as client. Passive objects are one, which passively waits for the message to be processed. It waits for another object that requires its services. In simple words it can be referred as server.
73. What is meant by software development method?
Software development method describes how to model and build software systems in a reliable and reproducible way. To put it simple, methods that are used to represent ones' thinking using graphical notations.
==
C++ and OOPS Sample Question
74. What are models and meta models?
Model:
It is a complete description of something (i.e. system).
Meta model:
It describes the model elements, syntax and semantics of the notation that allows their manipulation.
75. What do you meant by static and dynamic modeling?
Static modeling is used to specify structure of the objects that exist in the problem domain. These are expressed using class, object and USECASE diagrams.
But Dynamic modeling refers representing the object interactions during runtime. It is represented by sequence, activity, collaboration and statechart diagrams.
76. Why generalization is very strong?
Even though Generalization satisfies Structural, Interface, Behaviour properties. It is mathematically very strong, as it is Antisymmetric and Transitive.
Transitive: A=>B, B=>c then A=>c.
A. Salesman.
B. Employee.
C. Person.
Note:
All the other relationships satisfy all the properties like Structural properties, Interface properties, Behaviour properties.
77. Differentiate Aggregation and containment?
Aggregation is the relationship between the whole and a part. We can add/subtract some properties in the part (slave) side. It won't affect the whole part.
Best example is Car, which contains the wheels and some extra parts. Even though the parts are not there we can call it as car.
But, in the case of containment the whole part is affected when the part within that got affected. The human body is an apt example for this relationship. When the whole body dies the parts (heart etc) are died.
78. Can link and Association applied interchangeably?
No, You cannot apply the link and Association interchangeably. Since link is used represent the relationship between the two objects.
But Association is used represent the relationship between the two classes.
link :: student:Abhilash course:MCA
Association:: student course
79. what is meant by "method-wars"?
Before 1994 there were different methodologies like Rumbaugh, Booch, Jacobson, Meyer etc who followed their own notations to model the systems. The developers were in a dilemma to choose the method which best accomplishes their needs. This particular span was called as "method-wars"
80. Whether unified method and unified modeling language are same or different?
Unified method is convergence of the Rumbaugh and Booch.
Unified modeling lang. is the fusion of Rumbaugh, Booch and Jacobson as well as Betrand Meyer (whose contribution is "sequence diagram"). Its' the superset of all the methodologies.
===============
unix
Unix Sample Questions
Questions on file management in uinx
Following are some unix sample questions.
1. How are devices represented in UNIX?
Answer:
All devices are represented by files called special files that are located in/dev directory. Thus, device files and other files are named and accessed in the same way. A 'regular file' is just an ordinary data file in the disk. A 'block special file' represents a device with characteristics similar to a disk (data transfer in terms of blocks). A 'character special file' represents a device with characteristics similar to a keyboard (data transfer is by stream of bits in sequential order).
2. What is 'inode'?
Answer:
All UNIX files have its description stored in a structure called 'inode'. The inode contains info about the file-size, its location, time of last access, time of last modification, permission and so on. Directories are alsorepresented as files and have an associated inode. In addition to descriptions about the file, the inode contains pointers to the data blocks of the file. If the file is large, inode has indirect pointer to a block of pointers to additional data blocks (this further aggregates for larger files). A block is typically 8k.
Inode consists of the following fields:
* File owner identifier
* File type
* File access permissions
* File access times
* Number of links
* File size
* Location of the file data
3. Brief about the directory representation in UNIX
Answer:
A Unix directory is a file containing a correspondence between filenames and inodes. A directory is a special file that the kernel maintains. Only kernel modifies directories, but processes can read directories. The contents of adirectory are a list of filename and inode number pairs. When new directories are created, kernel makes two entries named '.' (refers to thedirectory itself) and '..' (refers to parent directory). System call for creating directory is mkdir (pathname, mode).
4. What are the Unix system calls for I/O?
Answer:
* open(pathname,flag,mode) - open file
* creat(pathname,mode) - create file
* close(filedes) - close an open file
* read(filedes,buffer,bytes) - read data from an open file
* write(filedes,buffer,bytes) - write data to an open file
* lseek(filedes,offset,from) - position an open file
* dup(filedes) - duplicate an existing file descriptor
* dup2(oldfd,newfd) - duplicate to a desired file descriptor
* fcntl(filedes,cmd,arg) - change properties of an open file
* ioctl(filedes,request,arg) - change the behaviour of an open file
The difference between fcntl anf ioctl is that the former is intended for any open file, while the latter is for device-specific operations.
5. How do you change File Access Permissions?
Answer:
Every file has following attributes:
* owner's user ID ( 16 bit integer )
* owner's group ID ( 16 bit integer )
* File access mode word
'r w x -r w x- r w x'
(user permission-group permission-others permission)
r-read, w-write, x-execute
To change the access mode, we use chmod(filename,mode).
Example 1:
To change mode of myfile to 'rw-rw-r--' (ie. read, write permission for user - read,write permission for group - only read permission for others) we give the args as:
chmod(myfile,0664) .
Each operation is represented by discrete values
'r' is 4
'w' is 2
'x' is 1
Therefore, for 'rw' the value is 6(4+2).
Example 2:
To change mode of myfile to 'rwxr--r--' we give the args as:
chmod(myfile,0744).
==
Unix Sample Questions
Questions on file management in uinx
6. What are links and symbolic links in UNIX file system?
Answer:
A link is a second name (not a file) for a file. Links can be used to assign more than one name to a file, but cannot be used to assign a directory more than one name or link filenames on different computers.
Symbolic link 'is' a file that only contains the name of another file.Operation on the symbolic link is directed to the file pointed by the it.Both the limitations of links are eliminated in symbolic links.
Commands for linking files are:
Link ln filename1 filename2
Symbolic link ln -s filename1 filename2
7. What is a FIFO?
Answer:
FIFO are otherwise called as 'named pipes'. FIFO (first-in-first-out) is a special file which is said to be data transient. Once data is read from named pipe, it cannot be read again. Also, data can be read only in the order written. It is used in interprocess communication where a process writes to one end of the pipe (producer) and the other reads from the other end (consumer).
8. How do you create special files like named pipes and device files?
Answer:
The system call mknod creates special files in the following sequence.
1. kernel assigns new inode,
2. sets the file type to indicate that the file is a pipe, directory or special file,
3. If it is a device file, it makes the other entries like major, minor device numbers.
For example:
If the device is a disk, major device number refers to the disk controller and minor device number is the disk.
9. Discuss the mount and unmount system calls
Answer:
The privileged mount system call is used to attach a file system to a directory of another file system; the unmount system call detaches a file system. When you mount another file system on to your directory, you are essentially splicing one directory tree onto a branch in another directory tree. The first argument to mount call is the mount point, that is , a directory in the current file naming system. The second argument is thefile system to mount to that point. When you insert a cdrom to your unix system's drive, the file system in the cdrom automatically mounts to /dev/cdrom in your system.
10. How does the inode map to data block of a file?
Answer:
Inode has 13 block addresses. The first 10 are direct block addresses of the first 10 data blocks in the file. The 11th address points to a one-level index block. The 12th address points to a two-level (double in-direction) index block. The 13th address points to a three-level(triple in-direction)index block. This provides a very large maximum file size with efficient access to large files, but also small files are accessed directly in one disk read.
11. What is a shell?
Answer:
A shell is an interactive user interface to an operating system services that allows an user to enter commands as character strings or through a graphical user interface. The shell converts them to system calls to the OS or forks off a process to execute the command. System call results and other information from the OS are presented to the user through an interactive interface. Commonly used shells are sh,csh,ks etc.
==
Unix Sample Questions
Questions on Process model and IPC in uinx
12. Brief about the initial process sequence while the system boots up.
Answer:
While booting, special process called the 'swapper' or 'scheduler' is created with Process-ID 0. The swapper manages memory allocation for processes and influences CPU allocation.
The swapper inturn creates 3 children:
the process dispatcher,
vhand and
dbflush
with IDs 1,2 and 3 respectively.
This is done by executing the file /etc/init. Process dispatcher gives birth to the shell. Unix keeps track of all the processes in an internal data structure called the Process Table (listing command is ps -el).
13. What are various IDs associated with a process?
Answer:
Unix identifies each process with a unique integer called ProcessID. The process that executes the request for creation of a process is called the 'parent process' whose PID is 'Parent Process ID'. Every process is associated with a particular user called the 'owner' who has privileges over the process. The identification for the user is 'UserID'. Owner is the user who executes the process. Process also has 'Effective User ID' which determines the access privileges for accessing resources like files.
getpid() -process id
getppid() -parent process id
getuid() -user id
geteuid() -effective user id
14. Explain fork() system call.
Answer:
The 'fork()' used to create a new process from an existing process. The new process is called the child process, and the existing process is called the parent. We can tell which is which by checking the return value from 'fork()'. The parent gets the child's pid returned to him, but the child gets 0 returned to him.
15. Predict the output of the following program code
main()
{
fork();
printf("Hello World!");
}
Answer:
Hello World!Hello World!
Explanation:
The fork creates a child that is a duplicate of the parent process. The child begins from the fork().All the statements after the call to fork() will be executed twice.(once by the parent process and other by child). The statement before fork() is executed only by the parent process.
16. Predict the output of the following program code
main()
{
fork(); fork(); fork();
printf("Hello World!");
}
Answer:
"Hello World" will be printed 8 times.
Explanation:
2^n times where n is the number of calls to fork()
==
Unix Sample Questions
Questions on Process model and IPC in uinx
17. List the system calls used for process management:
Answer:
System calls Description
fork() To create a new process
exec() To execute a new program in a process
wait() To wait until a created process completes its execution
exit() To exit from a process execution
getpid() To get a process identifier of the current process
getppid() To get parent process identifier
nice() To bias the existing priority of a process
brk() To increase/decrease the data segment size of a process
18. How can you get/set an environment variable from a program?
Answer:
Getting the value of an environment variable is done by using 'getenv()'.
Setting the value of an environment variable is done by using 'putenv()'.
19. How can a parent and child process communicate?
Answer:
A parent and child can communicate through any of the normal inter-process communication schemes (pipes, sockets, message queues, shared memory), but also have some special ways to communicate that take advantage of their relationship as aparent and child. One of the most obvious is that the parent can get the exit status of the child.
20. What is a zombie?
Answer:
When a program forks and the child finishes before the parent, the kernel still keeps some of its information about the child in case the parent might need it - for example, the parent may need to check the child's exit status. To be able to get this information, the parent calls 'wait()'; In the interval between the child terminating and the parent calling 'wait()', the child is said to be a 'zombie' (If you do 'ps', the child will have a 'Z' in its status field to indicate this.)
21. What are the process states in Unix?
Answer:
As a process executes it changes state according to its circumstances. Unix processes have the following states:
Running : The process is either running or it is ready to run .
Waiting : The process is waiting for an event or for a resource.
Stopped : The process has been stopped, usually by receiving a signal.
Zombie : The process is dead but have not been removed from the process table.
22. What Happens when you execute a program?
Answer:
When you execute a program on your UNIX system, the system creates a special environment for that program. This environment contains everything needed for the system to run the program as if no other program were running on the system. Each process has process context, which is everything that is unique about the state of the program you are currently running. Every time you execute a program the UNIX system does a fork, which performs a series of operations to create a process context and then execute your program in that context. The steps include the following: Allocate a slot in the process table, a list of currently running programs kept by UNIX. Assign a unique process identifier (PID) to the process. iCopy the context of the parent, the process that requested the spawning of the new process. Return the new PID to the parent process. This enables the parent process to examine or control the process directly.
After the fork is complete, UNIX runs your program.
23. What Happens when you execute a command?
Answer:
When you enter 'ls' command to look at the contents of your current working directory, UNIX does a series of things to create an environment for ls and the run it: The shell has UNIX perform a fork. This creates a new process that the shell will use to run the ls program. The shell has UNIX perform an exec of the ls program. This replaces the shell program and data with the program and data for ls and then starts running that new program. The ls program is loaded into thenew process context, replacing the text and data of the shell. The ls program performs its task, listing the contents of the current directory.
===
Unix Sample Questions
Questions on Process model and IPC in uinx
24. What is a Daemon?
Answer:
A daemon is a process that detaches itself from the terminal and runs, disconnected, in the background, waiting for requests and responding to them. It can also be defined as the background process that does not belong to a terminal session. Many system functions are commonly performed by daemons, including the sendmail daemon, which handles mail, and the NNTP daemon, which handles USENET news. Many other daemons may exist. Some of the most common daemons are: init: Takes over the basic running of the system when the kernel has finished the boot process. inetd: Responsible for starting network services that do not have their own stand-alone daemons. For example, inetd usually takes care of incoming rlogin, telnet, and ftp connections. cron: Responsible for running repetitive tasks on a regular schedule.
25. What is 'ps' command for?
Answer:
The ps command prints the process status for some or all of the running processes. The information given are the process identification number (PID),the amount of time that the process has taken to execute so far etc.
26. How would you kill a process?
Answer:
The kill command takes the PID as one argument; this identifies which process to terminate. The PID of a process can be got using 'ps' command.
27. What is an advantage of executing a process in background?
Answer:
The most common reason to put a process in the background is to allow you to do something else interactively without waiting for the process to complete. At the end of the command you add the special background symbol, &. This symbol tells your shell to execute the given command in the background.
Example:
cp *.* ../backup& (cp is for copy)
28. How do you execute one program from within another?
Answer:
The system calls used for low-level process creation are execlp() and execvp(). The execlp call overlays the existing program with the new one , runs that and exits. The original program gets back control only when an error occurs.
execlp(path,file_name,arguments..); //last argument must be NULL
A variant of execlp called execvp is used when the number of arguments is not known in advance. execvp(path,argument_array); //argument array should be terminated by NULL
29. What is IPC? What are the various schemes available?
Answer:
The term IPC (Inter-Process Communication) describes various ways by which different process running on some operating system communicate between each other. Various schemes available are as follows:
Pipes:
One-way communication scheme through which different process can communicate. The problem is that the two processes should have a common ancestor (parent-child relationship). However this problem was fixed with the introduction of named-pipes (FIFO).
Message Queues :
Message queues can be used between related and unrelated processes running on a machine.
Shared Memory:
This is the fastest of all IPC schemes. The memory to be shared is mapped into the address space of the processes (that are sharing). The speed achieved is attributed to the fact that there is no kernel involvement. But this scheme needs synchronization.
Various forms of synchronisation are mutexes, condition-variables, read-write locks, record-locks, and semaphores.
===
Unix Sample Questions
Questions on memory management in uinx
30. What is the difference between Swapping and Paging?
Answer:
Swapping:
Whole process is moved from the swap device to the main memory for execution. Process size must be less than or equal to the available main memory. It is easier to implementation and overhead to the system. Swapping systems does not handle the memory more flexibly as compared to the paging systems.
Paging:
Only the required memory pages are moved to main memory from the swap device for execution. Process size does not matter. Gives the concept of the virtual memory. It provides greater flexibility in mapping the virtual address space into the physical memory of the machine. Allows more number of processes to fit in themain memory simultaneously. Allows the greater process size than the available physical memory. Demand paging systems handle the memory more flexibly.
31. What is major difference between the Historic Unix and the new BSD release of Unix System V in terms of Memory Management?
Answer:
Historic Unix uses Swapping – entire process is transferred to the main memory from the swap device, whereas the Unix System V uses Demand Paging – only the part of the process is moved to the main memory. Historic Unix uses one Swap Device and Unix System V allow multiple Swap Devices.
32. What is the main goal of the Memory Management?
Answer:
It decides which process should reside in the main memory, Manages the parts of the virtual address space of a process which is non-core resident, Monitors the available main memory and periodically write the processes into the swap device to provide more processes fit in the main memory simultaneously.
33. What is a Map?
Answer:
A Map is an Array, which contains the addresses of the free space in the swap device that are allocatable resources, and the number of the resource units available there. This allows First-Fit allocation of contiguous blocks of a resource. Initially the Map contains one entry – address (block offset from the starting of the swap area) and the total number of resources.
Kernel treats each unit of Map as a group of disk blocks. On the allocation and freeing of the resources Kernel updates the Map for accurate information.
34. What scheme does the Kernel in Unix System V follow while choosing a swap device among the multiple swap devices?
Answer:
Kernel follows Round Robin scheme choosing a swap device among the multiple swap devices in Unix System V.
35. What is a Region?
Answer:
A Region is a continuous area of a process's address space (such as text, data and stack). The kernel in a 'Region Table' that is local to the process maintains region. Regions are sharable among the process.
36. What are the events done by the Kernel after a process is being swapped out from the main memory?
Answer:
When Kernel swaps the process out of the primary memory, it performs the following:
* Kernel decrements the Reference Count of each region of the process. If the reference count becomes zero, swaps the region out of themain memory,
* Kernel allocates the space for the swapping process in the swap device,
* Kernel locks the other swapping process while the current swapping operation is going on,
* The Kernel saves the swap address of the region in the region table.
===
Unix Sample Questions
Questions on memory management in uinx
37. Is the Process before and after the swap are the same? Give reason.
Answer:
Process before swapping is residing in the primary memory in its original form. The regions (text, data and stack) may not be occupied fully by the process, there may be few empty slots in any of the regions and whileswapping Kernel do not bother about the empty slots while swapping the process out.
After swapping the process resides in the swap (secondary memory) device. The regions swapped out will be present but only the occupied region slots but not the empty slots that were present before assigning.
While swapping the process once again into the main memory, the Kernel referring to the Process Memory Map, it assigns the main memory accordingly taking care of the empty slots in the regions.
38. What do you mean by u-area (user area) or u-block?
Answer:
This contains the private data that is manipulated only by the Kernel. This is local to the Process, i.e. each process is allocated a u-area.
39. What are the entities that are swapped out of the main memory while swapping the process out of the main memory?
Answer:
All memory space occupied by the process, process's u-area, and Kernel stack are swapped out, theoretically.
Practically, if the process's u-area contains the Address Translation Tables for the process then Kernel implementations do not swap the u-area.
40. What is Fork swap?
Answer:
fork() is a system call to create a child process. When the parent process calls fork() system call, the child process is created and if there is short ofmemory then the child process is sent to the read-to-run state in the swap device, and return to the user state without swapping the parent process. When the memory will be available the child process will be swapped into the main memory.
41. What is Expansion swap?
Answer:
At the time when any process requires more memory than it is currently allocated, the Kernel performs Expansion swap. To do this Kernel reserves enough space in the swap device. Then the address translation mapping is adjusted for the new virtual address space but the physicalmemory is not allocated. At last Kernel swaps the process into the assigned space in the swap device. Later when the Kernel swaps the process into the main memory this assigns memory according to the new address translation mapping.
==
Unix Sample Questions
Questions on memory management in uinx
42. How the Swapper works?
Answer:
The swapper is the only process that swaps the processes. The Swapper operates only in the Kernel mode and it does not uses System calls instead it uses internal Kernel functions for swapping. It is the archetype of all kernel process.
43. What are the processes that are not bothered by the swapper? Give Reason.
Answer:
Zombie process: They do not take any up physical memory.
Processes locked in memories that are updating the region of the process. Kernel swaps only the sleeping processes rather than the 'ready-to-run' processes, as they have the higher probability of being scheduled than the Sleeping processes.
44. What are the requirements for a swapper to work?
Answer:
The swapper works on the highest scheduling priority. Firstly it will look for any sleeping process, if not found then it will look for the ready-to-run process for swapping. But the major requirement for the swapper to work the ready-to-run process must be core-resident for at least 2 seconds before swapping out. And for swapping in the process must have been resided in the swap device for at least 2 seconds. If the requirement is not satisfied then the swapper will go into the wait state on that event and it is awaken once in a second by theKernel.
45. What are the criteria for choosing a process for swapping into memory from the swap device?
Answer:
The resident time of the processes in the swap device, the priority of the processes and the amount of time the processes had been swapped out.
46. What are the criteria for choosing a process for swapping out of the memory to the swap device?
Answer:
The process's memory resident time,
Priority of the process and
The nice value.
47. What do you mean by nice value?
Answer:
Nice value is the value that controls {increments or decrements} the priority of the process. This value that is returned by the nice () system call. The equation for using nice value is:
Priority = ("recent CPU usage"/constant) + (base- priority) + (nice value)
Only the administrator can supply the nice value. The nice () system call works for the running process only. Nice value of one process cannot affect the nice value of the other process.
48. What are conditions on which deadlock can occur while swapping the processes?
Answer:
All processes in the main memory are asleep.
All 'ready-to-run' processes are swapped out.
There is no space in the swap device for the new incoming process that are swapped out of the main memory.
There is no space in the main memory for the new incoming process.
==
Unix Sample Questions
Questions on memory management in uinx
49. What are conditions for a machine to support Demand Paging?
Answer:
Memory architecture must based on Pages,
The machine must support the 'restartable' instructions.
50. What is 'the principle of locality'?
Answer:
It's the nature of the processes that they refer only to the small subset of the total data space of the process. i.e. the process frequently calls the same subroutines or executes the loop instructions.
51. What is the working set of a process?
Answer:
The set of pages that are referred by the process in the last 'n', references, where 'n' is called the window of the working set of the process.
52. What is the window of the working set of a process?
Answer:
The window of the working set of a process is the total number in which the process had referred the set of pages in the working set of the process.
53. What is called a page fault?
Answer:
Page fault is referred to the situation when the process addresses a page in the working set of the process but the process fails to locate the page in the working set. And on a page fault the kernel updates the working set by reading the page from the secondary device.
54. What are data structures that are used for Demand Paging?
Answer:
Kernel contains 4 data structures for Demand paging. They are,
* Page table entries,
* Disk block descriptors,
* Page frame data table (pfdata),
* Swap-use table.
55. What are the bits that support the demand paging?
Answer:
Valid, Reference, Modify, Copy on write, Age. These bits are the part of the page table entry, which includes physical address of the page and protection bits.
Page address Age Copy on write Modify Reference Valid Protection
==
Unix Sample Questions
Questions on memory management in uinx
56. How the Kernel handles the fork() system call in traditional Unix and in the System V Unix, while swapping?
Answer:
Kernel in traditional Unix, makes the duplicate copy of the parent's address space and attaches it to the child's process, while swapping. Kernel in System V Unix, manipulates the region tables, page table, and pfdata table entries, by incrementing the reference count of the region table of shared regions.
57. Difference between the fork() and vfork() system call?
Answer:
During the fork() system call the Kernel makes a copy of the parent process's address space and attaches it to the child process.
But the vfork() system call do not makes any copy of the parent's address space, so it is faster than the fork() system call. The child process as a result of the vfork() system call executes exec() system call. The child process from vfork() system call executes in the parent'saddress space (this can overwrite the parent's data and stack ) which suspends the parent process until the child process exits.
58. What is BSS(Block Started by Symbol)?
Answer:
A data representation at the machine level, that has initial values when a program starts and tells about how much space the kernel allocates for the un-initialized data. Kernel initializes it to zero at run-time.
59. What is Page-Stealer process?
Answer:
This is the Kernel process that makes rooms for the incoming pages, by swapping the memory pages that are not the part of the working set of a process. Page-Stealer is created by the Kernel at the system initialization and invokes it throughout the lifetime of the system. Kernel locks a region when a process faults on a page in the region, so that page stealer cannot steal the page, which is being faulted in.
60. Name two paging states for a page in memory?
Answer:
The two paging states are:
The page is aging and is not yet eligible for swapping,
The page is eligible for swapping but not yet eligible for reassignment to other virtual address space.
61. What are the phases of swapping a page from the memory?
Answer:
Page stealer finds the page eligible for swapping and places the page number in the list of pages to be swapped.
Kernel copies the page to a swap device when necessary and clears the valid bit in the page table entry, decrements the pfdata reference count, and places the pfdata table entry at the end of the free list if its reference count is 0.
==
Unix Sample Questions
Questions on memory management in uinx
62. What is page fault? Its types?
Answer:
Page fault refers to the situation of not having a page in the main memory when any process references it.
There are two types of page fault :
Validity fault,
Protection fault.
63. In what way the Fault Handlers and the Interrupt handlers are different?
Answer:
Fault handlers are also an interrupt handler with an exception that the interrupt handlers cannot sleep. Fault handlers sleep in the context of theprocess that caused the memory fault. The fault refers to the running process and no arbitrary processes are put to sleep.
64. What is validity fault?
Answer:
If a process referring a page in the main memory whose valid bit is not set, it results in validity fault.
The valid bit is not set for those pages: that are outside the virtual address space of a process, that are the part of the virtual address space of the process but no physical address is assigned to it.
65. What does the swapping system do if it identifies the illegal page for swapping?
Answer:
If the disk block descriptor does not contain any record of the faulted page, then this causes the attempted memory reference is invalid and the kernel sends a "Segmentation violation" signal to the offendingprocess. This happens when the swapping system identifies any invalid memory reference.
66. What are states that the page can be in, after causing a page fault?
Answer:
On a swap device and not in memory,
On the free page list in the main memory,
In an executable file,
Marked "demand zero",
Marked "demand fill".
67. In what way the validity fault handler concludes?
Answer:
It sets the valid bit of the page by clearing the modify bit.
It recalculates the process priority.
==
Unix Sample Questions
Questions on memory management in uinx
68. At what mode the fault handler executes?
Answer:
At the Kernel Mode.
69. What do you mean by the protection fault?
Answer:
Protection fault refers to the process accessing the pages, which do not have the access permission. A process also incur the protection fault when it attempts to write a page whose copy on write bit was set during the fork() system call.
70. How the Kernel handles the copy on write bit of a page, when the bit is set?
Answer:
In situations like, where the copy on write bit of a page is set and that page is shared by more than one process, the Kernel allocates new page and copies the content to the new page and the other processes retain their references to the old page. After copying theKernel updates the page table entry with the new page number. Then Kernel decrements the reference count of the old pfdata table entry.
In cases like, where the copy on write bit is set and no processes are sharing the page, the Kernel allows the physical page to be reused by the processes. By doing so, it clears the copy on write bit and disassociates the page from its disk copy (if one exists), because other process may share the disk copy. Then it removes the pfdata table entry from the page-queue as the new copy of the virtual page is not on the swap device. It decrements the swap-use count for the page and if count drops to 0, frees the swap space.
71. For which kind of fault the page is checked first?
Answer:
The page is first checked for the validity fault, as soon as it is found that the page is invalid (valid bit is clear), the validity faulthandler returns immediately, and the process incur the validity page fault. Kernel handles the validity fault and the process will incur the protection fault if any one is present.
72. In what way the protection fault handler concludes?
Answer:
After finishing the execution of the fault handler, it sets the modify and protection bits and clears the copy on write bit. It recalculates the process-priority and checks for signals.
73. How the Kernel handles both the page stealer and the fault handler?
Answer:
The page stealer and the fault handler thrash because of the shortage of the memory. If the sum of the working sets of all processes is greater that the physical memory then the faulthandler will usually sleep because it cannot allocate pages for a process. This results in the reduction of the system throughput becauseKernel spends too much time in overhead, rearranging the memory in the frantic pace.
=====
os
Operating Systems Sample Questions
1. Explain the concept of Reentrancy.
Answer:
It is a useful, memory-saving technique for multiprogrammed timesharing systems. A Reentrant Procedure is one in which multiple users can share a single copy of a program during the same period. Reentrancy has 2 key aspects: The program code cannot modify itself, and the local data for each user process must be stored separately. Thus, the permanent part is the code, and the temporary part is the pointer back to the calling program and local variables used by that program. Each execution instance is called activation. It executes the code in the permanent part, but has its own copy of local variables/parameters. The temporary part associated with each activation is the activation record. Generally, the activation record is kept on the stack. Note: A reentrant procedure can be interrupted and called by an interrupting program, and still execute correctly on returning to the procedure.
2. Explain Belady's Anomaly.
Answer:
Also called FIFO anomaly. Usually, on increasing the number of frames allocated to a process' virtual memory, the process execution is faster, because fewer page faults occur. Sometimes, the reverse happens, i.e., the execution time increases even when more frames are allocated to the process. This is Belady's Anomaly. This is true for certain page reference patterns.
3. What is a binary semaphore? What is its use?
Answer:
A binary semaphore is one, which takes only 0 and 1 as values. They are used to implement mutual exclusion and synchronize concurrent processes.
4. What is thrashing?
Answer:
It is a phenomenon in virtual memory schemes when the processor spends most of its time swapping pages, rather than executing instructions. This is due to an inordinate number of page faults.
5. List the Coffman's conditions that lead to a deadlock.
Answer:
* Mutual Exclusion: Only one process may use a critical resource at a time.
* Hold & Wait: A process may be allocated some resources while waiting for others.
* No Pre-emption: No resource can be forcible removed from a process holding it.
* Circular Wait: A closed chain of processes exist such that each process holds at least one resource needed by another process in the chain.
6. What are short-, long- and medium-term scheduling?
Answer:
Long term scheduler determines which programs are admitted to the system for processing. It controls the degree of multiprogramming. Once admitted, a job becomes a process. Medium term scheduling is part of the swapping function. This relates to processes that are in a blocked or suspended state. They are swapped out of real-memory until they are ready to execute. The swapping-in decision is based on memory-management criteria.
Short term scheduler, also know as a dispatcher executes most frequently, and makes the finest-grained decision of which process should execute next. This scheduler is invoked whenever an event occurs. It may lead to interruption of one process by preemption.
7. What are turnaround time and response time?
Answer:
Turnaround time is the interval between the submission of a job and its completion. Response time is the interval between submission of a request, and the first response to that request.
8. What are the typical elements of a process image?
Answer:
* User data: Modifiable part of user space. May include program data, user stack area, and
* programs that may be modified.
* User program: The instructions to be executed.
* System Stack: Each process has one or more LIFO stacks associated with it. Used to store
* parameters and calling addresses for procedure and system calls.
* Process control Block (PCB): Info needed by the OS to control processes.
9. What is the Translation Lookaside Buffer (TLB)?
Answer:
In a cached system, the base addresses of the last few referenced pages is maintained in registers called the TLB that aids in faster lookup. TLB contains those page-table entries that have been most recently used. Normally, each virtual memory reference causes 2 physical memory accesses-- one to fetch appropriate page-table entry, and one to fetch the desired data. Using TLB in-between, this is reduced to just one physical memory access in cases of TLB-hit.
10. What is the resident set and working set of a process?
Answer:
Resident set is that portion of the process image that is actually in real-memory at a particular instant. Working set is that subset of resident set that is actually needed for execution. (Relate this to the variable-window size method for swapping techniques.)
===
Operating Systems Sample Questions
11. When is a system in safe state?
Answer:
The set of dispatchable processes is in a safe state if there exists at least one temporal order in which all processes can be run to completion without resulting in a deadlock.
12. What is cycle stealing?
Answer:
We encounter cycle stealing in the context of Direct Memory Access (DMA). Either the DMA controller can use the data bus when the CPU does not need it, or it may force the CPU to temporarily suspend operation. The latter technique is calledcycle stealing. Note that cycle stealing can be done only at specific break points in an instruction cycle.
13. What is meant by arm-stickiness?
Answer:
If one or a few processes have a high access rate to data on one track of a storage disk, then they may monopolize the device by repeated requests to that track. This generally happens with most common device scheduling algorithms (LIFO, SSTF, C-SCAN, etc). High-density multisurface disks are more likely to be affected by this than low density ones.
14. What are the stipulations of C2 level security?
Answer:
* C2 level security provides for: Discretionary Access Control
* Identification and Authentication
* Auditing
* Resource reuse
15. What is busy waiting?
Answer:
The repeated execution of a loop of code while waiting for an event to occur is called busy-waiting. The CPU is not engaged in any real productive activity during this period, and the process does not progress toward completion.
16. Explain the popular multiprocessor thread-scheduling strategies.
Answer:
* Load Sharing: Processes are not assigned to a particular processor. A global queue of threads is maintained. Each processor, when idle, selects a thread from this queue. Note that load balancing refers to a scheme where work is allocated to processors on a more permanent basis.
* Gang Scheduling: A set of related threads is scheduled to run on a set of processors at the same time, on a 1-to-1 basis. Closely related threads / processes may be scheduled this way to reduce synchronization blocking, and minimize process switching. Group scheduling predated this strategy.
* Dedicated processor assignment: Provides implicit scheduling defined by assignment of threads to processors. For the duration of program execution, each program is allocated a set of processors equal in number to the number of threads in the program. Processors are chosen from the available pool.
* Dynamic scheduling: The number of thread in a program can be altered during the course of execution.
17. When does the condition 'rendezvous' arise?
Answer:
In message passing, it is the condition in which, both, the sender and receiver are blocked until the message is delivered.
18. What is a trap and trapdoor?
Answer:
Trapdoor is a secret undocumented entry point into a program used to grant access without normal methods of access authentication. A trap is a software interrupt, usually the result of an error condition.
19. What are local and global page replacements?
Answer:
Local replacement means that an incoming page is brought in only to the relevant process' address space. Global replacement policy allows any page frame from any process to be replaced. The latter is applicable to variable partitions model only.
20. Define latency, transfer and seek time with respect to disk I/O.
Answer:
Seek time is the time required to move the disk arm to the required track. Rotational delay or latency is the time it takes for the beginning of the required sector to reach the head. Sum of seek time (if any) and latency is the access time. Time taken to actually transfer a span of data is transfer time.
==
Operating Systems Sample Questions
21. Describe the Buddy system of memory allocation.
Answer:
Free memory is maintained in linked lists, each of equal sized blocks. Any such block is of size 2^k. When some memory is required by a process, the block size of next higher order is chosen, and broken into two. Note that the two such pieces differ in address only in their kth bit. Such pieces are called buddies. When any used block is freed, the OS checks to see if its buddy is also free. If so, it is rejoined, and put into the original free-block linked-list.
22. What is time-stamping?
Answer:
It is a technique proposed by Lamport, used to order events in a distributed system without the use of clocks. This scheme is intended to order events consisting of the transmission of messages. Each system 'i' in the network maintains a counter Ci. Every time a system transmits a message, it increments its counter by 1 and attaches the time-stamp Ti to the message. When a message is received, the receiving system 'j' sets its counter Cj to 1 more than the maximum of its current value and the incoming time-stamp Ti. At each site, the ordering of messages is determined by the following rules: For messages x from site i and y from site j, x precedes y if one of the following conditions holds....(a) if Ti
23. How are the wait/signal operations for monitor different from those for semaphores?
Answer:
If a process in a monitor signal and no task is waiting on the condition variable, the signal is lost. So this allows easier program design. Whereas in semaphores, every operation affects the value of the semaphore, so the wait and signal operations should be perfectly balanced in the program.
24. In the context of memory management, what are placement and replacement algorithms?
Answer:
Placement algorithms determine where in available real-memory to load a program. Common methods are first-fit, next-fit, best-fit. Replacement algorithms are used when memory is full, and one process (or part of a process) needs to be swapped out to accommodate a new program. The replacement algorithm determines which are the partitions to be swapped out.
25. In loading programs into memory, what is the difference between load-time dynamic linking and run-time dynamic linking?
Answer:
For load-time dynamic linking: Load module to be loaded is read into memory. Any reference to a target external module causes that module to be loaded and the references are updated to a relative address from the start base address of the application module. With run-time dynamic loading: Some of the linking is postponed until actual reference during execution. Then the correct module is loaded and linked.
26. What are demand- and pre-paging?
Answer:
With demand paging, a page is brought into memory only when a location on that page is actually referenced during execution. With pre-paging, pages other than the one demanded by a page fault are brought in. The selection of such pages is done based on common access patterns, especially for secondary memory devices.
27. Paging a memory management function, while multiprogramming a processor management function, are the two interdependent?
Answer:
Yes.
28. What is page cannibalizing?
Answer:
Page swapping or page replacements are called page cannibalizing.
29. What has triggered the need for multitasking in PCs?
Answer:
Increased speed and memory capacity of microprocessors together with the support fir virtual memory and Growth of client server computing
30. What are the four layers that Windows NT have in order to achieve independence?
Answer:
* Hardware abstraction layer
* Kernel
* Subsystems
* System Services.
===
Operating Systems Sample Questions
31. What is SMP?
Answer:
To achieve maximum efficiency and reliability a mode of operation known as symmetric multiprocessing is used. In essence, with SMP any process or threads can be assigned to any processor.
32. What are the key object oriented concepts used by Windows NT?
Answer:
Encapsulation
Object class and instance
33. Is Windows NT a full blown object oriented operating system? Give reasons.
Answer:
No Windows NT is not so, because its not implemented in object oriented language and the data structures reside within one executive component and are not represented as objects and it does not support object oriented capabilities .
34. What is a drawback of MVT?
Answer:
It does not have the features like
* ability to support multiple processors
* virtual storage
* source level debugging
35. What is process spawning?
Answer:
When the OS at the explicit request of another process creates a process, this action is called process spawning.
36. How many jobs can be run concurrently on MVT?
Answer:
15 jobs
37. List out some reasons for process termination.
Answer:
* Normal completion
* Time limit exceeded
* Memory unavailable
* Bounds violation
* Protection error
* Arithmetic error
* Time overrun
* I/O failure
* Invalid instruction
* Privileged instruction
* Data misuse
* Operator or OS intervention
* Parent termination.
38. What are the reasons for process suspension?
Answer:
* swapping
* interactive user request
* timing
* parent process request
39. What is process migration?
Answer:
It is the transfer of sufficient amount of the state of process from one machine to the target machine
40. What is mutant?
Answer:
In Windows NT a mutant provides kernel mode or user mode mutual exclusion with the notion of ownership.
==
Operating Systems Sample Questions
41. What is an idle thread?
Answer:
The special thread a dispatcher will execute when no ready thread is found.
42. What is FtDisk?
Answer:
It is a fault tolerance disk driver for Windows NT.
43. What are the possible threads a thread can have?
Answer:
* Ready
* Standby
* Running
* Waiting
* Transition
* Terminated.
44. What are rings in Windows NT?
Answer:
Windows NT uses protection mechanism called rings provides by the process to implement separation between the user mode and kernel mode.
45. What is Executive in Windows NT?
Answer:
In Windows NT, executive refers to the operating system code that runs in kernel mode.
46. What are the sub-components of I/O manager in Windows NT?
Answer:
* Network redirector/ Server
* Cache manager.
* File systems
* Network driver
* Device driver
47. What are DDks? Name an operating system that includes this feature.
Answer:
DDks are device driver kits, which are equivalent to SDKs for writing device drivers. Windows NT includes DDks.
48. What level of security does Windows NT meets?
Answer:
C2 level security.
====
Data Structure Sample Questions
A data structure is a way of storing data in a computer so that it can be used efficiently.
Different kinds of data structures are suited to different kinds of applications.
Following are some Data Structure Sample questions:
1. What is data structure?
Answer: A data structure is a way of organizing data that considers not only the items stored, but also their relationship to each other. Advance knowledge about the relationship between data items allows designing of efficient algorithms for the manipulation of data.
2. List out the areas in which data structures are applied extensively?
Answer: The name of areas are:
* Compiler Design,
* Operating System,
* Database Management System,
* Statistical analysis package,
* Numerical Analysis,
* Graphics,
* Artificial Intelligence,
* Simulation
3. What are the major data structures used in the following areas : RDBMS, Network data model & Hierarchical data model.
Answer: The major data structures used are as follows:
* RDBMS - Array (i.e. Array of structures)
* Network data model - Graph
* Hierarchical data model - Trees
4. If you are using C language to implement the heterogeneous linked list, what pointer type will you use?
Answer: The heterogeneous linked list contains different data types in its nodes and we need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So we go for void pointer. Void pointer is capable of storing pointer to any type as it is a generic pointer type.
5. Minimum number of queues needed to implement the priority queue?
Answer: Two. One queue is used for actual storing of data and another for storing priorities.
6. What is the data structures used to perform recursion?
Answer: Stack. Because of its LIFO (Last In First Out) property it remembers its 'caller' so knows whom to return when the function has to return. Recursion makes use of system stack for storing the return addresses of the function calls.
Every recursive function has its equivalent iterative (non-recursive) function. Even when such equivalent iterative procedures are written, explicit stack is to be used.
7. What are the notations used in Evaluation of Arithmetic Expressions using prefix and postfix forms?
Answer: Polish and Reverse Polish notations.
==
Data Structure Sample Questions
8. Convert the expression ((A + B) * C - (D - E) ^ (F + G)) to equivalent Prefix and Postfix notations.
Answer: Prefix Notation: ^ - * +ABC - DE + FG
Postfix Notation: AB + C * DE - - FG + ^
9. How many null branches are there in a binary tree with 20 nodes?
Answer: 21
Let us take a tree with 5 nodes (n=5)
It will have only 6 (ie,5+1) null branches.
A binary tree with n nodes has exactly n+1 null nodes.
10. What are the methods available in storing sequential files?
Answer: The methods available in storing sequential files are:
* Straight merging,
* Natural merging,
* Polyphase sort,
* Distribution of Initial runs.
11. How many different trees are possible with 10 nodes ?
Answer: 1014
For example, consider a tree with 3 nodes(n=3), it will have the maximum combination of 5 different (ie, 23 - 3 = 5) trees.
i ii iii iv v
In general:
If there are n nodes, there exist 2n-n different trees.
12. List out few of the Application of tree data-structure?
Answer: The list is as follows:
* The manipulation of Arithmetic expression,
* Symbol Table construction,
* Syntax analysis.
13. List out few of the applications that make use of Multilinked Structures?
Answer: The applications are listed below:
* Sparse matrix,
* Index generation.
14. In tree construction which is the suitable efficient data structure?
Answer: Linked list is the efficient data structure.
15. What is the type of the algorithm used in solving the 8 Queens problem?
Answer: Backtracking
Data Structure Sample Questions
16. In an AVL tree, at what condition the balancing is to be done?
Answer: If the 'pivotal value' (or the 'Height factor') is greater than 1 or less than -1.
17. What is the bucket size, when the overlapping and collision occur at same time?
Answer: One. If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.
18. Traverse the given tree using Inorder, Preorder and Postorder traversals.
Answer:
* Inorder : D H B E A F C I G J
* Preorder: A B D H E C F G I J
* Postorder: H D E B F I J G C A
19. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could have formed a full binary tree?
Answer: 15.
In general:
There are 2n-1 nodes in a full binary tree.
By the method of elimination:
Full binary trees contain odd number of nodes. So there cannot be full binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete binary tree but not a full binary tree. So the correct answer is 15.
20. In the given binary tree, using array you can store the node 4 at which location?
Answer: At location 6
1 2 3 - - 4 - - 5
Root LC1 RC1 LC2 RC2 LC3 RC3 LC4 RC4
where LCn means Left Child of node n and RCn means Right Child of node n
21. Sort the given values using Quick Sort?
65 70 75 80 85 60 55 50 45
Answer:
Sorting takes place from the pivot value, which is the first value of the given elements, this is marked bold. The values at the left pointer and right pointer are indicated using L and R respectively.
65 70L 75 80 85 60 55 50 45R
Since pivot is not yet changed the same process is continued after interchanging the values at L and R positions
65 45 75L 80 85 60 55 50R 70
65 45 50 80L 85 60 55R 75 70
65 45 50 55 85L 60R 80 75 70
65 45 50 55 60R 85L 80 75 70
When the L and R pointers cross each other the pivot value is interchanged with the value at right pointer. If the pivot is changed it means that the pivot has occupied its original position in the sorted order (shown in bold italics) and hence two different arrays are formed, one from start of the original array to the pivot position-1 and the other from pivot position+1 to end.
60L 45 50 55R 65 85L 80 75 70R
55L5 45 50R 60 65 70R 80L 75 85
50L 45R 55 60 65 70 80L 75R 85
In the next pass we get the sorted form of the array.
45 50 55 60 65 70 75 80 85
==
Data Structure Sample Questions
22. For the given graph, draw the DFS and BFS?
Answer:
* BFS: A X G H P E M Y J
* DFS: A X H P E Y M J G
23. Classify the Hashing Functions based on the various methods by which the key value is found.
Answer: The list of Hashing functions is as follows:
* Direct method
* Subtraction method
* Modulo-Division method
* Digit-Extraction method
* Mid-Square method
* Folding method
* Pseudo-random method
24. What are the types of Collision Resolution Techniques and the methods used in each of the type?
Answer: The types of Collision Resolution Techniques are:
* Open addressing (closed hashing)
The methods used include:
o Overflow block
* Closed addressing (open hashing)
The methods used include:
o Linked list
o Binary tree
25. In RDBMS, what is the efficient data structure used in the internal storage representation?
Answer: B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes searching easier. This corresponds to the records that shall be stored in leaf nodes.
26. Draw the B-tree of order 3 created by inserting the following data arriving in sequence - 92 24 6 7 11 8 22 4 5 16 19 20 78
Answer:
27. What is a spanning Tree?
Answer: A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.
28. Does the minimum spanning tree of a graph give the shortest distance between any 2 specified nodes?
Answer: No. Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn't mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.
29. Convert the given graph with weighted edges to minimal spanning tree.
Answer: the equivalent minimal spanning tree is:
==
Data Structure Sample Questions
30. Whether Linked List is linear or Non-linear data structure?
Answer: According to Access strategies Linked list is a linear one.
According to Storage Linked List is a Non-linear one.
31. Draw a binary Tree for the expression : A * B - (C + D) * (P / Q)
Answer:
32. For the following COBOL code, draw the Binary tree?
01 STUDENT_REC.
02 NAME.
03 FIRST_NAME PIC X(10).
03 LAST_NAME PIC X(10).
02 YEAR_OF_STUDY.
03 FIRST_SEM PIC XX.
03 SECOND_SEM PIC XX.
Answer:
« Previous
=========
ccne
CCNA Sample Questions
640-801
The Building Scalable Cisco Certification exam is a qualifying exam for the CCNA certifications. The exam will certify that the successful candidate has important knowledge and skills necessary to use advanced IP addressing and routing in implementing scalability for Cisco routers connected to LANs and WANs. The exam covers topics on Advanced IP Addressing, Routing Principles, Configuring the EIGRP, Configuring the Open Shortest Path First Protocol, Configuring IS-IS, Manipulating Routing Updates, and configuring basic BGP.
We provide CCNA sample questions that help test your preparedness for taking the actual cisco certification exam. CCNA 640-801 sample questions help in consolidating your concepts, preparation and also as exam cram. Each CCNA sample question is based on respective vendors published exam objectives and designed to help attaincisco certification.
1. Which of the following about Routing Protocols is correct? (choose all that apply)
1. RIP 2 is a Distance Vector Protocol
2. The RIP 1 uses bandwidth and delay as the metric
3. RIP 1 uses subnet mask
4. The holdown timer value for RIP is 240 sec
5. The Update timer value of IGRP is 90 sec
Answer :A & E
2. What command is used to delete the configuration stored in NVRAM?
1. erase startup-config
2. erase running-config
3. delete nvram
4. erase nvram
Answer :A
3. What is the maximum number of hops that OSPF allows before marking a network as unreachable?
1. 15
2. 16
3. 255
4. Unlimited
5. 99
Answer :D
4. Which routing protocol would allow a network administrator scalability, VLSM support and minimize overhead if the network administrator wants to merge different networks all using routers from multiple vendors?
1. VTP
2. RIP
3. IGRP
4. EIGRP
5. OSPF
Answer :E
5. Two routers Rtr1 and Rtr2 are both configured with RIP only. What will be the result when Rtr1 receives a routing update that contains a higher cost path to a network already in its routing table?
1. The update information will replace the existing routing table entry.
2. The update information will be added to the existing routing table.
3. The existing routing table entry will be deleted from the routing table and all routers will exchange routing updates to reach convergence.
4. The update will be ignored and no further action will be taken.
Answer :E
6. Two routers Rtr1 and Rtr2 are both configured with RIP only. What will be the result when Rtr1 receives a routing update that contains a higher cost path to a network already in its routing table?
1. The update information will replace the existing routing table entry.
2. The update information will be added to the existing routing table.
3. The existing routing table entry will be deleted from the routing table and all routers will exchange routing updates to reach convergence.
4. The update will be ignored and no further action will be taken.
Answer :E
7. What switching method examines the destination MAC address as the frame is being received and then begins forwarding the frame prior to receiving the entire frame?
1. Modified Cut Through
2. Store and Forward
3. Cut Through
4. Fragment Free
Answer :C
==
#
3. U" interface
4. a standard pc
5. an ISDN terminal adapter
Answer :A & D
# Which protocol is used to look up an IP address from a known Ethernet address?
1. TCP
2. RARP
3. ARP
4. IP
Answer :B
# You need to troubleshoot a Cisco router at the Toronto office of How2Pass Inc. The router loses its configuration each time it is rebooted. You study the output displayed in the exhibit. What is the cause of the problem?
Exhibit:
----- output omitted ----
Cisco 2620 (MPC860) processor (revision 0x200) with 16384/2048K bytes of memory.
Processor board ID JAD05076EF6 (3878188963)
M860 processor: part number 0, mask 49
Bridging software.
X.25 software, Version 3.0.0.
2 Ethernet/IEEE 802.3 interface(s)
2 Serial(sync/async) network interface(s)
2 Low speed serial(sync/async) network interface(s)
32K bytes of non-volatile configuration memory.
16384K bytes of processor board System flash (Read/Write)
Configuration register is 0x2142
1. There is insufficient RAM for the IOS image
2. The configuration register is misconfigured
3. There is insufficient flash memory
4. NVRAM failed POST
5. There is insufficient NVRAM
B
# If an Ethernet port on router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?
1. 2046
2. 8190
3. 4094
4. 4096
5. 1024
Answer :C
# The Frame Relay circuit between router A and router B is experiencing congestion. Which type of notification are used to alleviate the congestion? (Choose three)
1. DE
2. BECN
3. DLCI 100 is Down
4. CIR
5. FECN
Answer :D, B & A
# At which OSI layer does data translation and code formatting occur?
1. Network
2. Physical
3. Data link
4. Transport
5. Session
6. Presentation
Answer :F
# Which one of the following is a reason to use a hardware address?
1. To transmit a packet from one local device to another local device.
2. To transmit a frame from one interface to another interface.
3. To obtain a vendor code/serial number from the user.
4. To transmit data from one local device to remote device across Internet.
5. To contain logical information about a device to use an end-to-end transmission.
Answer :B
==
CCNA Sample Questions
640-801
15. Which wireless data communication type has a high data rate but is limited to very short distances?
1. Infrared
2. Broadband Personal Communication Service (PCS)
3. Narrowband
4. Spread spectrum
Answer :A
16. Ethernet networks are broadcast domains and collision domains. How the hosts on an Ethernet network will know when to resume transmissions after a collision has occurred? (Choose all that apply)
1. The router on the segment will signal that the collision has cleared.
2. The jam signal indicates that the collision has been cleared.
3. The hosts will attempt to resume transmission after a time delay has expired.
4. The destination host sends a request to the source for retransmission.
5. An electrical pulse indicates that the collision has cleared.
Answer :C
17. Which command must be entered when connecting two routers without external DCE devices via a serial link?
1. Serial up.
2. Dte rate.
3. Dce rate.
4. Clock rate.
5. Line protocol up.
Answer :D
18. How2Pass Router Simulation. Please read the instructions and problem statement carefully. Complete your configuration and press the Done button when finished.
Answer :
Troubleshooting
In the simulation, click Host D, a dialog box will appear, select the Cisco Terminal option and press the OK button, to establish a terminal session to the router Kenan.
Press ‹Enter› key to enter the User EXEC mode.
Enter the privileged EXEC mode by typing enable at the Kenan> prompt and then the password how2pass at the Password: prompt. The privileged EXEC mode prompt, Kenan#, will be displayed.
Issue the command show running-config (abbreviated as sh runn), read the output carefully to find any missing or wrong configuration. Whenever you see the prompt --More-- at the bottom of a page, you can press spacebar to see the next output page.
Notice the running-config entries for interface serial1. It has wrong IP address and subnet mask. Notice the word shutdown below Serial1.
To confirm this, issue show interface serial 1 command. The output confirms that the interface is shutdown.
This simulation also supports ping, show ip protocol and show ip route commands for effective debugging.
Do the configuration shown below and ping to a remote ip address.
Save your configuration using copy running-config startup-config (abbreviated as copy runn start) command.
Now press the Done button.
Configuration Required
Kenan#config t
Enter configuration commands, one per line. End with CNTL/Z.
Kenan(config)#int s1
Kenan(config-if)#ip address 192.168.69.2 255.255.255.0
Kenan(config-if)#no shutdown
%LINEPROTO-5-UPDOWN: Line protocol on Interface Serial1, changed state to up
%LINK-3-UPDOWN: Interface Serial1 changed state to up
Kenan(config-if)#^z
%SYS-5-CONFIG_I: Configured from console by console
Kenan#ping 192.168.59.1
Type escape sequence to abort.
Sending 5, 100-byte ICMP Echos to 192.168.69.1, timeout is 2 seconds: ! ! ! ! !
Success rate is 100 percent (5/5), round-trip min/avg/max = 4/4/4 ms
Kenan#copy runn start
Destination filename [startup-config]?
Building configuration...
19. Which of the following will configure a static route on Router A to network 180.18.30.0/24 with an administrative distance of 90?
1. Router(config)# ip route 90 180.18.30.0 255.255.255.0 182.18.20.2
2. Router(config)# ip route 180.18.30.0 255.255.255.0 182.18.20.2 90
3. Router(config)# ip route 180.18.30.1 255.255.255.0 182.18.20.1 90
4. Router(config)# ip route 180.18.20.1 255.255.255.0 182.18.30.0 90
5. Router(config)# ip route 90 180.18.20.1 255.255.255.0 182.18.20.2
Answer :B
20. What is a trunk link?
1. A link that is only part of one VLAN and is referred to as the native VLAN of the port
2. A link that can carry multiple VLANs
3. A switch port connected to the Internet
4. Data and voice capability on the same interface
Answer :B
==
CCNA Sample Questions
640-801
21. What does the concept route aggregation mean when one talks about using variable subnet masking?
1. Combining routes to multiple networks into one supernet.
2. Deleting unusable addresses through the creation of many subnets.
3. Calculating the available host addresses in the AS.
4. Reclaiming unused space by means of changing the subnet size.
Answer :A
22. What PPP protocol provides dynamic addressing, authentication, and multilink?
1. HDLC
2. LCP
3. NCP
4. X.25
Answer :B
23. Study the network topology exhibit carefully, in particular the two switches SW1, SW2, and the router RT3. Which statements are true in this scenario? (Select two)
1. The hosts on the 192.168.1.0 network form one collision domain, and the hosts on the 192.168.2.0 network form a second collision domain.
2. Each host is in a separate collision domain.
3. All the devices in both networks will receive a broadcast to 255.255.255.255 sent by host A.
4. Only the devices in network 192.168.1.0 will receive a broadcast to 255.255.255.255 sent by host A.
5. All the devices on both networks are members of the same collision domain.
Answer :D & B
24. What is the purpose of Inverse ARP?
1. It is used to map a known DLCI to an IP address
2. It is used to map a known DLCI to a MAC address
3. It is used to map a known IP address to a MAC address
4. It is used to map a known MAC address to an IP address
5. It is used to map a known MAC address to DLCI
Answer :A
25. You want to configure a router for load balancing across 4 unequal cost paths on your network. Which of the following routing protocols can you use? (Choose two)
1. RIP v1
2. RIP v2
3. OSPF
4. IGRP
5. EIGRP
6. VLSM
Answer :D & E
26. At what layer data is split into segments
1. Transport
2. LAN
3. Session
4. Data Link
Answer :A
27. Host A is communicating with the server. What will be the source MAC address of the frames received by Host A from the server?
1. The MAC address of Host A.
2. The MAC address of the server network interface.
3. The MAC address of router interface E1.
4. The MAC address of router interface E0.
Answer :D
28. What switching method examines the destination MAC address as the frame is being received and then begins forwarding the frame prior to receiving the entire frame?
1. Fragment Free
2. Store and Forward
3. Modified Cut Through
4. Cut Through
Answer :D
==
CCNA Sample Questions
640-801
29. The exhibit displays the partial contents of an encapsulation header. Which of the following are true of the network traffic represented in this diagram? (Select three)
1. This is traffic from an FTP server.
2. This is a UDP header.
3. The last PDU received in this session had a sequence number of 292735.
4. This is an OSI layer 4 header.
5. This is traffic from a Telnet client.
Answer :D, A &C
30. What does the concept route aggregation mean when one talks about using variable subnet masking?
1. Reclaiming unused space by means of changing the subnet size.
2. Combining routes to multiple networks into one supernet.
3. Calculating the available host addresses in the AS.
4. Deleting unusable addresses through the creation of many subnets.
Answer :B
31. What OSI layer is FRAME-RELAY mapped to?
1. Network
2. Transport
3. Data Link
4. Physical
Answer :C
32. Please drag the appropriate options to the correct targets. After completing the question press the push button below the question to save your response.
Answer: 1:E,2:A,3:F
Explanation:
LAPD - provides the data link protocol that allows delivery of messages across that D-channel to the local switch.
LAPB - Protocol and is designed primarily to satisfy the signaling requirements of ISDN basic access. It is defined by ITU-T Recommendations Q.920 and Q.921.
TE1 - ISDN -capable four-wire cable. Understands signaling and 2B=D. Uses an S reference point.
ITU.T.430 - Defines connectors, encoding, framing, and reference points.
TE2 - Equipment that does not understand ISDN protocols and specifications (no ISDN awareness).
Uses an R reference point, typically an RS-232 or V.35 cable, to connect to a TA.
NT1 - CPE equipment in North America. Connects with a U reference point (two-wire) to the telco.
33. Which routing protocol would allow a network administrator scalability, VLSM support and minimize overhead if the network administrator wants to merge different networks all using routers from multiple vendors?
1. OSPF
2. VTP
3. EIGRP
4. RIP
5. IGRP
Answer :A
34. What Channel is provided by ISDN BRI?
1. 2B+1D
2. 30B+1D
3. 23B+1D
4. 2D+1B
Answer :A
35. Which of the following describe the process identifier that is used to run OSPF on a router? (Choose two)
1. It is globally significant.
2. It is an optional parameter required only if multiple OSPF processes are running on the router.
3. All routers in the same OSPF area must have the same process ID if they are to exchange routing information.
4. It is needed to identify a unique instance of an OSPF database.
5. It is locally significant.
Answer :D & E
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CCNA Sample Questions
640-801
36. Given the the following network diagram, assume that port 1 through 3 are assigned to VLAN 1 and ports 4 through 6 are assigned to VLAN 2 on each switch. The switches are interconnected over a trunked link.
Which of the following conditions would verify VLAN and trunk operation? (choose 3)
1. A. Host 1-1 can ping Host 1-2
2. B. Host 1-1 can ping Host 4-2
3. C. Host 1-1 can not ping Host 1-2
4. D. Host 4-1 can not ping Host 1-2
5. E. Host 4-1 can ping Host 4-2
Answer :A, D, E
37. By looking at the configuration, Which additional command must be issued on the Branch router before interesting traffic will be sent to the Remote router?
Hostname: Branch Hostname: Remote
PH# 123-6000, 123-6001 PH# 123-8000, 123-8001
SPID1: 32055512360001 SPID1: 32055512380001
SPID2: 32055512360002 SPID2: 32055512380002
isdn switch-type basic ni
username Remote password cisco
interface bri0
ip address 10.1.1.1 255.255.255.0
encapsulation ppp
ppp authentication chap
isdn spid1 41055512360001
isdn spid2 41055512360002
dialer map ip 10.1.1.2 name Remote 1238001
dialer-list 1 protocol ip permit
1. (config-if)# dialer-group 1
2. (config-if)# dialer-list 1
3. (config-if)# dialer map 1
4. (config-if)# dialer-route 1
Answer : A
Explanation :The "dialer-group #" command tells the access-list (used with the dialer-list # command), which interface to activate when it finds interesting traffic. The numbers at end of each command must match.
38. You are the network administrator of the RouterSim global software Company. You receive a call from a user who is unable to reach a server at a remote site. After further review you discover the following info:
Local PC 190.0.3.35/24
Default Gateway 190.0.3.1
Remote Server 190.0.5.250/24
You then conduct the following tests from the offending local PC: Ping 127.0.0.1 - Unsuccessful
Ping 190.0.3.35 - Successful
Ping 190.0.3.1 - Unsuccessful
Ping 190.0.5.250 - Unsuccessful
Which of the following problems would create the test results listed above?
1. TCP/IP not correctly installed
2. Local physical layer problem
3. NIC not functioning
4. Remote physical layer problem
Answer : A
Explanation :If you cannot ping the loopback address of 127.0.0.1, then something is wrong with the IP configuration of the host.
39. Choose three reasons why the networking industry uses a layered model. (Choose 3)
1. Allow changes in one layer to occur without changing other layers
2. To get Gigabit speeds on LANs
3. Clarify what general functions is to be done rather than how to do it
4. To create filter tables on LANs
5. Order network troubleshooting steps
Answer :A, C, E
Explanation :
The reasons to use a layer model are:
1. Clarify what general function is to be done rather than how to do it.
2. Reduce the complexity of networking into more manageable sublayers.
3. Enable interoperability using standard interfaces.
4. Allow changes in one layer to occur without changing other layers.
5. Speed up network industry progress by allowing specialization.
6. Allow for shortcut explanations to facilitate protocol comparisons.
7. Order network troubleshooting steps.
8. Facilitate systematic troubleshooting.
40. What is a disadvantage to using bridges in your network?
1. Filters by MAC address
2. Stops broadcast storms
3. Doesn't stop broadcast storms
4. Can only use up to 4 bridges in any LAN
Answer : C
Explanation :
Even though the 5-4-3 rule specifies you can only have 4 bridges in a network, that is not really a disadvantage. The disadvantage to bridging is that it filters by MAC address and cannot create separate networks like a router can. This means that if a broadcast storm was to take place a bridge will forward the frames.
==
CCNA Sample Questions
640-801
41. You work as network administrator at Brain dump. Your trainee is configuring a router with both physical and logical interfaces. He asks you what factor determines the OSPF router ID. What should you tell him?
1. The lowest IP address of any interface.
2. The highest IP address of any interface.
3. The highest IP address of any logical interface.
4. The middle IP address of any logical interface.
5. The lowest IP address of any physical interface.
6. The highest IP address of any physical interface.
7. The lowest IP address of any logical interface.
Answer :F
Explanation: The OSPF topology database includes information about routers and the subnets, or links, to which they are attached. To identify the routers in the neighbor table’s topology database, OSPF uses a router ID (RID) for each router. A router’s OSPF RID is that router’s highest IP address on a physical interface when OSPF starts running.
Note: The OSPF router ID is a 32-bit IP address selected at the beginning of the OSPF process. Thehighest IP address configured on the router is the router ID. If a loopback address is configured, then it is the router ID. In case of multiple loopback addresses, the highest loopback address is the router ID. Once the router ID is elected it doesn't change unless the IP address is removed or OSPF restarts.
42. Which of the following routes will be used to forward data in a situation where a routing table contains static, RIP, and IGRP routes destined to the same network with each set to its default administrative distance?
1. The IGRP route
2. The static route
3. The RIP route
4. All three will load balance.
Answer :B
Explanation:To decide which route to use, IOS uses a concept called Administrative Distance. Administrative distance is a number that denotes how believable an entire routing protocol is on a single router. The lower the number, the better, or more believable the routing protocol.
Route Type Administrative Distance
* Static 1
* IGRP 100
* RIP 120
43. What is the basic characteristic of switches and hubs?
1. Hubs cannot filter frames.
2. Using hubs is costly with regard to bandwidth availability.
3. Switches do and can not forward broadcasts.
4. Switches are more efficient than hubs in processing frames.
5. Switches increase the number of collision domains in the network.
Answer :E
Explanation:Switches increases the number of collisions domains in the network. Switches that are configured with VLANs will reduce the size of the collision domains by increasing the number of collision domains in a network, but making them smaller than that of one big, flat network.
44. When you consider half-duplex and full-duplex Ethernet, what are unique for half-duplex Ethernet? (Select two options.)
1. Half-duplex Ethernet operates in a shared collision domain.
2. Half-duplex Ethernet operates in an exclusive broadcast domain.
3. Half-duplex Ethernet has efficient throughput.
4. Half-duplex Ethernet has lower effective throughput.
5. Half-duplex Ethernet operates in an exclusive collision domain.
Answer :A, D
Explanation:A single device could not be sending a frame and receiving a frame at the same time because it would mean that a collision was occurring. So, devices simply chose not to send a frame while receiving a frame. That logic is called half-duplex logic. Ethernet switches allow multiple frames to be sent over different ports at the same time. Additionally, if only one device is connected to a switch port, there is never a possibility that a collision could occur. So, LAN switches with only one device cabled to each port of the switch allow the use of full-duplex operation. Full duplex means that an Ethernet card can send and receive concurrently.
==
CCNA Sample Questions
640-801
45. You are a Cisco certified expert. You have been contracted by the Braindumps Pro chain to fix a problem that was caused by a MCP certified technician who could not complete the configuration of the routers. This Braindumps Pro chain has three stores and wanted to maintain their bicycle repair business in a centralized manner through network connectivity. They then asked the local MCP certified technician to configure the routers, but the technician failed to establish connectivity among the routers. The routers are named Braindumps1, Braindumps2, and Braindumps3, respectively. Identify the faults(s) and make the appropriate change(s) to rectify the configuration of the routers.
The MCP technician configured the routers with the specification that follows:
* The routers are named Brain dump1, Brain dump2, and Brain dump3.
* RIP is the routing protocol
* Clocking is provided on the serial 0 interfaces
* The password on each router is "Brain dump"
* The subnet mask on all interfaces is the default mask.
* The IP addresses are listed in chart below.
Brain dump1
E0 192.168.27.1
E1 192.168.29.1
S0 192.168.31.1
Secret password: Brain dump
Brain dump2
E0 192.168.35.1
S0 192.168.33.1
S1 192.168.31.2
Secret password: Brain dump
Brain dump3
E0 192.168.37.1
S1 192.168.33.2
Secret password: Brain dump
To configure the router click on the host icon that is connected to the router by a serial cable.
Answer :
Click on Host 2:
Router Brain dump1:
Brain dump1 enable
Password: Brain dump
Brain dump1 # config terminal
Brain dump1 (config) # interface ethernet 0
Brain dump1 (config-if) # ip address 192.168.27.1 255.255.255.0
Brain dump1 (config-if) # no shutdown
Brain dump1 (config-if) # exit
Brain dump1 (config) # interface ethernet 1
Brain dump1 (config-if) # ip address 192.168.29.1 255.255.255.0
Brain dump1 (config-if) # no shutdown
Brain dump1 (config-if) # exit
Brain dump1 (config) # interface serial 0
Brain dump1 (config-if) # ip address 192.168.31.1 255.255.255.0
Brain dump3 (config-if) # clock rate 64000
Brain dump1 (config-if) # no shutdown
Brain dump1 (config-if) # exit.
Brain dump1 (config) # router rip
Brain dump1 (config-router) # network 192.168.27.0
Brain dump1 (config-router) # network 192.168.29.0
Brain dump1 (config-router) # network 192.168-31.0
Brain dump1 (config-router) # Ctrl-Z
Brain dump1 # copy running-config startup-config
Click on Host 4
Router Brain dump2:
Brain dump2 enable
Password: Brain dump
Brain dump2 # config t
Brain dump2 (config) # interface ethernet 0
Brain dump2 (config-if) # ip address 192.168.35.1 255.255.255.0
Brain dump2 (config-if) # no shutdown
Brain dump2 (config-if) # exit
Brain dump2 (config) # interface serial 0
Brain dump2 (config-if) # ip address 192.168.33.1 255.255.255.0
Brain dump2 (config-if) # clock rate 64000
Brain dump2 (config-if) # no shutdown
Brain dump2 (config-if) # exit
Brain dump2 (config) # interface serial 1
Brain dump2 (config-if) # ip address 192.168.31.2 255.255.255.0
Brain dump2 (config-if) # no shutdown
Brain dump2 (config-if) # exit
Brain dump2 (config) # router rip
Brain dump2 (config-router) # network 192.168.35.0
Brain dump2 (config-router) # network 192.168.33.0
Brain dump2 (config-router) # network 192.168.31.0
Brain dump2 (config-router) # Ctrl-Z
Brain dump2 # copy running-config startup-config
Router Brain dump3:
Click on Host6
Brain dump3 enable
Password: Brain dump
Brain dump3 # config t
Brain dump3 (config) # interface ethernet 0
Brain dump3 (config-if) # ip address 192.168.37.1 255.255.255.0.
Brain dump3 (config-if) # no shutdown
Brain dump3 (config-if) # exit
Brain dump3 (config) # interface serial 1
Brain dump3 (config-if) # ip address 192.168.33.2 255.255.255.0
Brain dump3 (config-if) # no shutdown
Brain dump3 (config-if) # exit
Brain dump3 (config) # router rip
Brain dump3 (config-router) # network 192.168.33.0
Brain dump3 (config-router) # network 192.168.37.0
Brain dump3 (config-router) # Ctrl-Z
Brain dump3 # copy running-config startup-config
==
CCNA Sample Questions
640-801
46. The following exhibit shows the Braindumps.biz WAN. Study it carefully:
What are the broadcast addresses of the subnets in the Braindumps network? (Select three options).
A. 172.16.82.255
B. 172.16.95.255
C. 172.16.64.255
D. 172.16.32.255
E. 172.16.47.255
F. 172.16.79.255
Answer :B, E, F
Explanation:The subnets in the network are subnetted Class B addresses. A /20 subnet mask means that the subnet addresses increment by 16.
For example: 172.16.16.0, 172.16.32.0, 172.16.48.0, 172.16.64.0 etc. The broadcast address is the last IP address before the next subnet address.
B: The switch IP address (172.16.82.90) is in the 172.16.80.0 subnet. 172.16.95.255 is the broadcast address for the 172.16.80.0 subnet.
E: 172.16.47.255 is the broadcast address for the 172.16.32.0 subnet.
F: 172.16.79.255 is the broadcast address for the 172.16.64.0 subnet.
47. You are a network administrator at Braindumps. The Braindumps network is illustrated in the following exhibit. Study it carefully:
Routers Braindumps1 and Braindumps2 are connected through their social interfaces, however, they cannot communicate. You ascertain that Braindumps1 has the correct configuration.
Can you identify the fault on router Braindumps2?
1. Link reliability is insufficient
2. IPCP is not open
3. Incorrect subnet mask
4. Incompatible encapsulation
5. Bandwidth allocation is too low
6. Incomplete IP address
Answer :D
Explanation :HDLC and PPP Configuration
HDLC and PPP configuration is straightforward. You just need to be sure to configure the same WAN data-link protocol on each end of the serial link. Otherwise, the routers will misinterpret the incoming frames, because each WAN data-link protocol uses a different frame format. Other than configuring some optional features that are all you need to do.
==
CCNA Sample Questions
640-801
48. You are a network administrator at Braindumps. The Braindumps network is illustrated in the following exhibit. Study it carefully:
You want to prevent users on the Research Network and the Internet from accessing the Braindumps Support server, but you want to allow all other Braindumps users access to the server. You create an access control list (ACL) called research block. The ACL contains the following statements:
deny 172.16.102.0 0.0.0.255 172.16.104.255 0.0.0.0
permit 172.16.0.0 0.0.255.255 172.16.104.252 0.0.0.0
Which of the following commands sequences will place this list to meet these requirements?
1. Brain dump1 (config)# interface e0
Brain dump1 (config-if)# ip access-group research block in
2. Brain dump1 (config)# interface s0
Brain dump1 (config-if)# ip access-group research block out
3. Brain dump2 (config)# interface s0
Brain dump2 (config-if)# ip access-group research block out
4. Brain dump2 (config)# interface s1
Brain dump2 (config-if)# ip access-group research block in
5. Brain dump3 (config)# interface s1
Brain dump3 (config-if)# ip access-group research block in
6. Brain dump3 (config)# interface e0
Brain dump3 (config-if)# ip access-group research block out
Answer :F
Explanation:To enable the ACL on an interface and define the direction of packets to which the ACL is applied, the ip access-group command is used.
When referring to a router, these terms have the following meanings.
* Out - Traffic that has already been through the router and is leaving the interface; the source Would be where it's been (on the other side of the router) and the destination is where it's Going.
* In - Traffic that is arriving on the interface and which will go through the router; the source would Be where it's been and the destination is where it's going (on the other side of the router).
49. You are a network administrator at Braindumps. You are troubleshooting a router problem. You issue the show ip route command on one of the routers. The output from the command is shown in the following exhibit:
Router Brain dump# Show in ip route.
What does [120/3] represent?
1. 120 is the bandwidth allocation and 3 is the routing process number.
2. 120 is the administrative distance and 3 is the metric for that route.
3. 120 is the value of the update timer and 3 is the number of updates received.
4. 120 is the UDP port for forwarding traffic and 3 is the number of bridges.
Answer :B
Explanation:To decide which route to use, IOS uses a concept called Administrative Distance. Administrativedistance is a number that denotes how believable an entire routing protocol is on a single router. The lower the number, the better, or more believable the routing protocol.
Route Type Administrative Distance
Route Type Administrative Distance
* Connected 0
* EIGRP 90
* IGRP 100
* OSPF 110
* RIP 120
For RIP, the metric is the hop count, so in this case the route is 3 hops away.
==
CCNA Sample Questions
640-801
50. Which PPP authentication methods will you use when configuring PPP on an interface of a Cisco router? (Select two options.)
1. SSL
2. SLIP
3. PAP
4. LAPB
5. CHAP
6. VNP
Answer :C, E
Explanation :Password Authentication Protocol (PAP) and Challenge Handshake Authentication Protocol (CHAP) authenticate the endpoints on either end of a point-to-point serial link. Chap is the preferred method today because the identifying codes flowing over the link are created using a MD5 one-way hash, which is more secure that the clear-text passwords sent by PAP.
51. If NVRAM lacks boot system commands, where does the router look for the Cisco IOS by default?
1. ROM
2. RAM
3. Flash
4. Bootstrap
5. Startup-.config
Answer :C
Explanation :Flash memory - Either an EEPROM or a PCMCIA card, Flash memory stores fully functional IOS images and is the default location where therouter gets its IOS at boot time. Flash memory also can be used to store any other files, including configuration files.
52. What could be the rationale behind using passive-interface command when configuring a router?
1. Allows interfaces to share common IP addresses.
2. Allows an interface to remain up without the aid of keepalives.
3. Allows a router to send routing and not receive updates via that interface.
4. Allows a routing protocol to forward updates that is missing its IP address.
5. Allows a router to receive routing updates on an interface but not send updates via that interface.
Answer :E
Explanation:The passive-interface command is used to control the advertisement of routing information. The command enables the suppression of routing updates over some interfaces while allowing updates to be exchanged normally over other interfaces. For any interface specified as passive, no routing information will be sent. Routing information received on that interface will be accepted and processed by therouter. This is often useful for DDR links such as ISDN.
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==
CCNA Sample Questions
640-801
53. You are a systems administrator at Braindumps and you've just acquired a new Class C IP network. Which of one of the subnet masks below is capable of providing one useful subnet for each of the above departments (support, financial, sales & development) while still allowing enough usable host addresses for each departments needs?
1. 255.255.255.128
2. 255.255.255.192
3. 255.255.255.224
4. 255.255.255.240
5. 255.255.255.248
6. 255.255.255.252
Answer :C
Explanation:The network currently consists of 5 networks and another network will be acquired. That gives us atotal of 6 networks. This requires that we use 3 bits for the network address. Using the formula 2n-2 we get 6. This also leaves us with 5 bits for hosts, which gives us 30 hosts.
54. You are a network administrator at Braindumps. You need to troubleshoot the Braindumps network shown in the exhibit. Study the Exhibit carefully:
The host, EX1, is connected to the Braindumps1 LAN, but it cannot get access to resources on any ofthe other networks. The host's configuration is as follows:
Host address: 192.168.5.45
Subnet mask: 255.255.255.240
Default gateway: 192.168.5.32
Which of the following is the cause of this problem?
1. The default gateway is a subnetwork address.
2. The default gateway is on a different subnet address as the host.
3. The IP address of the host is on a different subnet.
4. The host subnet mask is incompatible to the subnet mask of the attached router interface.
Answer :A
Explanation:The range of the subnet used in this question is 192.168.166.32 to 192.168.166.47.192.168.166.32 is the network address and 192.168.166.47 is the broadcast. This leaving a usable host address range of 192.168.166.33 to 192.168.166.46.
The default gateway for the host should be 192.168.166.33.
55. Which of the following are benefits of segmenting a network with a router? (Select all that apply)
1. Broadcasts are not forwarded across the router.
2. All broadcasts are completely eliminated.
3. Adding a router to the network decreases latency.
4. Filtering can occur based on Layer 3 information.
5. Routers are more efficient than switches and will process the data more quickly.
6. None of the above.
Answer :A, D
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