带联合的 sizeof 结构 [英] sizeof struct with union

问题描述

我对整个数据对齐"的事情感到非常困惑:

#include int main(){结构体{int i,*p;字符 c;联合{字符类型[4];无符号字符 d;} un;};struct st s1={25,&a,'X',"asdf"};printf("s1 的大小是 %d",sizeof(s1));返回0；}

由于数据对齐，我认为由于

的大小

int i : 4 字节int *p : 8 字节字符 c : 1 字节 (+3)联合:4 个字节

输出将是 20，但这输出 sizeof s1 是 24！为什么输出24?这是否考虑到 int *p，即 8 个字节?

解决方案

在您的系统上，指针是 8 个字节并对齐到 8 个字节.

<前>1 2 3 4 5 6 7 8+---+---+---+---+---+---+---+---+---+|国际 |[垫] |+---+---+---+---+---+---+---+---+---[phone]+---+---+---+---+---+---+---+---+---+|整数 *p |+---+---+---+---+---+---+---+---+---[phone] 24+---+---+---+---+---+---+---+---+---+|| |联合国 |[垫] |+---+---+---+---+---+---+---+---+---+

当然，如果不了解 ABI，您就无法确定确切的布局.以上只是一个例子.

I'm very confused by the whole 'data alignment' thing:

#include <stdio.h>
int main(){
    struct st{
        int i,*p;
        char c;
        union { char type[4]; unsigned char d;} un; 
    };

    struct st s1={25,&a,'X',"asdf"};
    printf("sizeof s1 is %d",sizeof(s1));
    return 0;
}

due to data alignment, i thought that since the sizes of

int i : 4 bytes
int *p : 8 bytes
char c : 1 byte(+3)
union : 4 bytes

the output would be 20, but this outputs sizeof s1 is 24! Why is it outputting 24? Does this regard int *p, which is 8 bytes?

解决方案

On your system, pointers are 8 bytes and aligned to 8 bytes.

  1   2   3   4   5   6   7   8
+---+---+---+---+---+---+---+---+
| int i         | [pad]         |
+---+---+---+---+---+---+---+---+

  9  10  11  12  13  14  15  16
+---+---+---+---+---+---+---+---+
| int *p                        |
+---+---+---+---+---+---+---+---+

 17  18  19  20  21  22  23  24
+---+---+---+---+---+---+---+---+
| c | un            | [pad]     |
+---+---+---+---+---+---+---+---+

Of course, you cannot be sure of the exact layout without knowing the ABI. The above is only an example.

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