Linux内核---5.关于结构体对齐

1. 有一个结构体如下

#include <stdio.h>
#include <stdlib.h>
#pragma pack (4)

typedef struct  _STRU_TEST
{
    unsigned int unLen;
    unsigned char ucHeader;
    unsigned char ucType;
    unsigned short usAddr;
}STRU_TEST;

void main(void)
{
    STRU_TEST t1;
    int temp = sizeof(t1);
    printf("temp = %d\n",temp);
    return ;
}

打印出 temp=8,很完美。

2. 修改一下

typedef struct  _STRU_TEST
{
    unsigned int unLen;
    unsigned char ucHeader;
    unsigned short usAddr;
    unsigned char ucType;
}STRU_TEST;

打印出temp=12.

3. 只是调了一下顺序，为什么会出现这么大的差别呢？

两个规则：

a. 结构体内部: 编译器为结构的每个成员按其自然对界与#pragma  pack指定值中较小的那个进行对齐。

b. 结构体整体: 按照结构体中最大的数据成员和#pragma  pack指定值中较小的那个进行对齐。

typedef struct _STRU_TEST
{
    unsigned int unLen;  [0-3]
    unsigned char ucHeader; [4]
    unsigned char ucType; [5]
    unsigned short usAddr; [6-7]
}STRU_TEST1;
typedef struct _STRU_TEST
{
    unsigned int unLen;  [0-3]
    unsigned char ucHeader;  [4]
    unsigned short usAddr;  [6-7] //按照规则a,short类型自身长度为2, pragma为4，按照2字节对齐，所以这儿不是[5-6]而是[6-7]
    unsigned char ucType; [8]
}STRU_TEST2;

结构体STRU_TEST2实际占的内存空间为9个字节，但是结构体的对齐是按照规则b: 结构体最大数据成员int 为4字节，pragma中指定4，所以整个结构体按4字节对齐，所以sizeof=12.

4. linux 下指定对齐

gcc 有关键字 __attribute__来指定对齐

#include <stdio.h>
#include <stdlib.h>
#pragma pack (4)

typedef struct  _STRU_TEST
{
    unsigned int unLen;
    unsigned char ucHeader;
    unsigned short usAddr;
    unsigned char ucType;
//}STRU_TEST;
}__attribute__ ((packed, aligned(1)))STRU_TEST ;

void main(void)
{
    STRU_TEST t1;
    int temp = sizeof(t1);
    printf("temp = %d\n",temp);
    return ;
}

最后打印 temp=8; int 4+ char 1+ short 2+ char 1 = 8

[参考文章]http://www.cppblog.com/Tauruser/archive/2007/02/28/19049.html