1. Printf();函数中的参数是按从右到左顺序运算的
如：printf("%d %d\n", *ptr, *(++ptr)); 是先进行++ptr,再运行*ptr。
2. 为了便于结构体Struct内元素的访问和管理，当结构体内的元素的长度都小于处理器的位数时，便以结构体里最长的数据元素为对齐单位，也就是说，结构体的长度一定是最长的数据元素的整数倍。如果结构体内存长度大于处理器位数的元素，那么就以处理器的位数为对其单位。但是结构体内类型相同的连续元素将在连续的空间内。
数据对齐（data alignment）：是指数据所在的内存位置（内存地址）必须是该数据长度的整数倍。
3. 静态变量是存放在全局数据区中的，而sizeof计算的是栈中分配的大小，所以如果一个类中有static类型的变量，sizeof某个类时，该static变量是不计算在内的。
4. Class的sizeof大小的计算方式跟struct是一样的，也要考虑最大对齐单位。没有任何virtual元素（函数）的空类sizeof为1，单一继承、多继承空类的空类sizeof也为1，虚继承（Class B:public virtual A）的空类的sizeof为4。类中含有virtual函数（不管多少个virtual函数），该类的sizeof大小为该类去掉virtual的 sizeof大小再+4。
比如：
class A{
};
sizeof(A)为1。
class A{
int a;
public:
A(){}
~A(){}
virtual void f(){}
virtual int g(){}
};
sizeof(A)为8。
class A{
public:
A(){}
~A(){}
virtual void f(){}
virtual int g(){}
};
sizeof(A)为4。
class A{
public:
A(){}
~A(){}
void f(){}
int g(){}
};
sizeof(A)为1。
5. sizeof可以计算函数的大小，只是根据函数的返回类型的大小来计算的。比如 int f(int a,char b)，sizeof(f(a,b))结果为4。总之，对函数使用sizeof，在编译阶段会被函数返回值的类型取代。
6. sizeof()括号中的代码不会被编译的，所说int a=6; sizeof(a++); 后a的值仍然是6。
7. 只要是指针，sizeof返回的大小都是4。
8. sizeof求数组的大小返回的是各维数的乘机*数组元素的大小。
9. sizeof(string)为4.
10. int test(char var[])函数中，sizeof(var)为4，因为var[]等价于*var，已经退化为一个指针了。
11. Class B{
flaot f; char p; int adf[3];
}; 则cout<
12.内联函数(inline)被不包含for, while, switch语句
13. （1）void (*f) ( ) 函数指针；（2）void *f（）函数返回指针；（3）const int * const指针；（4）int * const 指向const的指针；（5）const int * const 指向const的const指针。
14. int a[]={1,2,3,4,5};
int * ptr=(int *)(&a+1);
printf("%d,%d",*(a+1), *(ptr-1));
输出结果为2，5。因为&a+1表示a+6，即ptr指向第六个位置。
牢记一点：数组名本身就是指针，再加个&，就变成了双指针，这里的双指针就是指二维指针，加1，就是数组整体加一行。
15. int *p; *p=3; //错误，因为p只是一个指针，*p并不存在。
16. 拷贝构造函数和拷贝复制函数的区别
17. 32位机器上，int和long 都是占4个字节；64位机器上，int和long分别占4、8个字节
18. int x=4; x+=x-=x-x--; 最后x为7.而在Java中相同的代码后x为8.
//test1
#include
int main()
{
int i=3;
int j;
void *k; k++; //compile error, no type
j=sizeof(++i + ++i); //type is int, so is 4B, not execute in sizeof()
printf("i= %d j= %d \n", i, j); //3 and 4
}
----------------------------------------------------------------------------
---------
//test2
#include
int main()
{
char *p = malloc(20); //sizeof(p)=4B/8B, sizeof(*p) = 1B
int a[5]={10, 20, 30, 40, 50}; /a and &a[0] is same pointstobase.&a+1-->whole array int b[5]={100, 200, 300, 400, 500}; //sizeof(b)-->4*5=20B
int *ptr = (int *)(&a+1); sizeof(a) is 5 elements, and p is an int pointer 4B
int *t = (int *)(&a -1);
printf("%d %d %d \n", *(a+1), *(ptr-1), *(t+1)); //20, 50, 200
//push to stack: 4G--50,40,30,20,10, 500,400,300,200,100,----0G
}
---------------------------------------------------------------------
//test3
#include
int main()
{
char *p;
char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8};
p = (buf+1)[5]; //buf+1-->2
printf("%d \n", p); //9
}
--------------------------------------------------------------------------
//test4
#include
int main()
{
int a[][3]={1,2,3,4,5,6}; // 123 456
int (*ptr)[3] = a; //int **x=a; compile wrong type
printf("%d %d \n", (*ptr)[1], (*ptr)[2]); //row0: col1 and 2: 2 3
++ptr;
printf("%d %d \n", (*ptr)[1], (*ptr)[2]); // row1: col1 2: 5 6
}
-----------------------------------------------------------------------
//test5
#include
void f(char **p);
int main()
{
char *argv[]={"ab", "cd", "ef", "gh"}; //argv[0]=ab, [1]=cd...
f(argv);
}
void f(char **p)
{
char *t;
t=(p+=sizeof(int))[-1]; p+=sizeof(int) -->back 1 [-1]-->gh
printf("%s\n", t); //gh
}
---------------------------------------------------------------------
//test6
#include
void foo(int b[][3]);
int main()
{
int a[3][3]={{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
foo(a); // must be same col
printf("%d\n", a[2][1]); // 8-->0
}
void foo(int b[][3])
{
++b; //b points 1,2,3 now points to 4 5 6 after ++b, b points 4 5 6
b[1][1]=0; // 8-->0
}
./test1
i= 3 j= 4
./test2
20 50 200
./test3
9
./test4
2 3
5 6
./test5
gh
./test6
0
/*
Method #1 (No tricks, just an array with empty first dimension)
===============================================================
You don't have to specify the first dimension!
*/
int func1(short mat[][3])
{
register short i, j;
printf(" Declare as matrix, explicitly specify second dimension: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
printf("%5.2d", mat[i][j]);
}
}
printf("\n");
return;
}
/*
Method #2 (pointer to array, second dimension is explicitly specified)
======================================================================
*/
int func2(short (*mat)[3])
{
register short i, j;
printf(" Declare as pointer to column, explicitly specify 2nd dim: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
printf("%5.2d", mat[i][j]);
}
}
printf("\n");
return;
}
/*
Method #3 (Using a single pointer, the array is "flattened")
============================================================
With this method you can create general-purpose routines.
The dimensions doesn't appear in any declaration, so you
can add them to the formal argument list.
The manual array indexing will probably slow down execution.
*/
int func3(short *mat)
{
register short i, j;
printf(" Declare as single-pointer, manual offset computation: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
printf("%5.2d", *(mat + 3*i + j));
}
}
printf("\n");
return;
}
/*
Method #4 (double pointer, using an auxiliary array of pointers)
================================================================
With this method you can create general-purpose routines,
if you allocate "index" at run-time.
Add the dimensions to the formal argument list.
*/
int func4(short **mat)
{
short i, j, *index[3];
for (i = 0 ; i < 3 ; i++)
index[i] = (short *)mat + 3*i;
printf(" Declare as double-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
printf("%5.2d", index[i][j]);
}
}
printf("\n");
return;
}
/*
Method #5 (single pointer, using an auxiliary array of pointers)
================================================================
*/
int func5(short *mat[3])
{
short i, j, *index[3];
for (i = 0 ; i < 3 ; i++)
index[i] = (short *)mat + 3*i;
printf(" Declare as single-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
printf("%5.2d", index[i][j]);
}
}
printf("\n");
return;
}
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