Memory alignment in C

Question

Here is a snippet:

#pragma pack(4)
struct s1
{
    char a;
    long b;
};
#pragma pack()

#pragma pack(2)
struct s2
{
    char c;
    struct s1 st1;
};
#pragma pack()


#pragma pack(2)
struct s3
{
    char a;
    long b;
};
#pragma pack()

#pragma pack(4)
struct s4
{
    char c;
    struct s3 st3;
};
#pragma pack()

I though sizeof(s4) should be 10 or 12. But it turns out to be 8. I am using Visual C++ 6.0. Could someone tell me why?

The reason why I thought sizeof(s4) is 10 based on:|c|Padding|Padding|Padding|a|Padding|b|b|b|b|. But the actual layout is |c|Padding|a|Padding|b|b|b|b|. — Ge Liu
– Ge Liu, Commented Nov 26, 2012 at 5:05
On your system, the sizeof(long) is 4. Thats why sizeof(s3) is 6 and sizeof(s4) is 8. — sgarizvi
– sgarizvi, Commented Nov 26, 2012 at 5:07

user541686 · Accepted Answer · 2012-11-26 05:17:31Z

2

#pragma pack(2)
struct s3
{
    char a;
    long b;
};
#pragma pack()

So the packing alignment of s3 is 2, and its size is 1 (alignment 1) + 1 (padding) + 4 (alignment 2) = 6.

#pragma pack(4)
struct s4
{
    char c;
    struct s3 st3;
};
#pragma pack()

The packing alignment of s4 is 4, and its size is 1 (alignment 1) + 1 (padding) + 6 (alignment 2) = 8.

Note that #pragma pack doesn't "extra-align" anything with looser alignment requirements. It only reduces alignment, i.e. controls "packing" alignment.

edited Nov 26, 2012 at 5:17

answered Nov 26, 2012 at 5:07

user541686

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4 Comments

Ge Liu Over a year ago

As the alignment of s4 is 4, It should be 1 + 3(padding) + 6 = 10. or 1 + 3(padding) + 6 + 2(padding) = 12.

user541686 Over a year ago

@user1758245: No. All it means is that each member of it will require alignment no greater than 4. It doesn't mean that each member will have alignment requirement 4. Notice, after all, it's called pack and not align for a reason -- precisely because it controls packing, not alignment. __declspec(align(#)) might be what you need though.

melpomene Over a year ago

user1758245: Why 3(padding)? The alignment of s3 is 2, so it only needs 1(padding) after char c.

Ge Liu Over a year ago

I finally understand "pack" and "align". Thank you. And thank all those who have answered my question.

Michael Burr · Accepted Answer · 2012-11-26 05:22:59Z

The documentation for #pragma pack(n) says that "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". However I think this is incorrect; the docs should say that the alignment of a member will be on a boundary that is either a multiple of n or the alignment requirement of the member, whichever is smaller.

The alignment requirement of struct s3 is 2 (due to the #pragma pack(2) in effect when it was declared). So it still gets an alignment of 2 even in the struct that has a #pragma pack(4) in effect. So the layout of struct s4 looks like:

 char c;
 char padding;      // for alignment of the `struct s3`
 char s3.a;
 char s3.padding;   // for alignment
 long s3.b;

Total size == 8.

Xymostech · Accepted Answer · 2012-11-26 05:05:07Z

0

In s3, char (1) + long (4) = 5 -> 6 (packed)

In s4, char (1) + struct s3 (6) = 7 -> 8 (packed)

I'm not actually sure how #pragma pack works, but this seems reasonable.

answered Nov 26, 2012 at 5:05

Xymostech

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Comments

antak · Accepted Answer · 2012-11-26 05:41:28Z

0

This verifies what Michael Burr said:

#include <stdio.h>

#pragma pack(push, 2)
struct s3
{
    long b;
};
#pragma pack(pop)

struct s4
{
    char c;
    struct s3 st3;
};

int main() {
    printf("%lu, %lu\n", sizeof(s3), sizeof(s4));
    return 0;
}

Output: 4, 6

answered Nov 26, 2012 at 5:41

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