Always same effect of #pragma pack(16) and #pragma pack(8)?

Question

I am trying to align data members by using #pragma pack (n). Take the following as an example:

#include <iostream>
using namespace std;

#pragma pack(8) // or (16)

struct A
{
    int a;
    char b;
    char c;
    char d;
    char e;
    char f;
    double g;
};
int main()
{
    cout << sizeof(A) << endl;

    return 0;
}

Both will print 24 for #pragma pack(8) and #pragma pack(16). I can understand the result for n=8 with the data alignment, of my understanding, as follows:

Bytes: |1 2 3 4|5|6|7|8|9|10 11 12 13 14 15 16|17 18 19 20 21 22 23 24|
 Data: |a      |b|c|d|e|f|padding             |g                      |

But I cannot understand why the result is still 24 for n=16. I also tried other examples, all of them seems to give same result for n=8 and n=16. Can someone explain why? Are the data members aligned the same way as n=8?

P.S.: Tested in VS2010 under Win-x64.

Cory Nelson · Accepted Answer · 2014-03-31T05:42:35.533

6

From the page you linked to:

The alignment of a member will be on a boundary that is either a multiple of n or a multiple of the size of the member, whichever is smaller.

For each member it takes the minimum alignment, between the member's optimal alignment and the pack value. The members in your example are aligned exactly the same regardless of pack(8) and pack(16) because the optimal alignment of the types are all less than 16.

If you had a member that required 16-byte alignment, such as __m128, you would be able to find different results.

edited Mar 31 '14 at 05:42

answered Mar 31 '14 at 05:36

Cory Nelson

29,236
5
72
110

Thanks. It makes sense now. Where can I find references for *`For each member it takes the minimum alignment, between the member's optimal alignment and the pack value.`*? – herohuyongtao Mar 31 '14 at 05:49
1

`pragma pack` is a compiler extension. I'm afraid the only reference will be the page you linked to. The part I quoted says essentially that. – Cory Nelson Mar 31 '14 at 05:52

score 3 · Answer 2 · answered Mar 31 '14 at 05:44

For pack to mean much, the struct in question must contain a member at least as large as the N you specify to the pragma. For example:

#include <iostream>
using namespace std;

#pragma pack(push)
#pragma pack(1)

struct A
{
    char a;
    double b;
};
#pragma pack(8)
struct B
{
    char a;
    double b;
};
#pragma pack(pop)

int main()
{
    cout << sizeof(A) << "\n" << sizeof(B) << endl;
    return 0;
}

On my computer, this prints:

9    
16

...which I'd consider fairly typical (at least for machines that accept pragma pack in the first place).

score 2 · Answer 3 · answered Mar 31 '14 at 05:36

Because pack specifies the maximum alignment used when needed. Values smaller then 16 bytes will still be aligned to smaller values because there is no gain in aligning them to larger values. SO an 32-bits integer is still aligned to 4 byes. A double is aligned to 8 bytes. SO you probably would see a difference when switching between 4 and 8.

Always same effect of #pragma pack(16) and #pragma pack(8)?

3 Answers3

Linked