sizeof struct with union [duplicate]

问题:

This question already has an answer here:

• How is the size of a C++ class determined? 4 answers

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?

回答1:

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.

回答2:

On the architecture you're using, the int *p is 8 bytes, but needs to be on an 8 byte boundary also. That gives you 4 bytes pad between i and p. Furthermore, because int *p needs to be on an 8 byte boundary, the overall structure needs to be a multiple of 8 bytes so that an array of them always has the right alignment for p.

So, you end up with a layout like this:

• i: 4 bytes, alignment 4
• pad: 4 bytes
• p: 8 bytes, alignment 8
• c: 1 bytes
• un: 4 bytes (as it can remain byte aligned)
• pad: 3 bytes (to round it out to an 8 byte boundary)