Why would the size of a packed structure be different on Linux and Windows when using gcc?

Question

In the code below, why is the size of the packed structure different on Linux and Windows when compiled with gcc?

#include <inttypes.h>
#include <cstdio>

// id3 header from an mp3 file
struct header
{    
        uint8_t version[ 2 ];
        uint8_t flags;
        uint32_t size;
} __attribute__((packed));

int main( int argc, char **argv )
{
        printf( "%u\n", (unsigned int)sizeof( header ) );
        return 0;
}

gcc versions used:

$ g++ --version
g++ (Ubuntu/Linaro 4.5.2-8ubuntu4) 4.5.2
$ x86_64-w64-mingw32-g++ --version
x86_64-w64-mingw32-g++ (GCC) 4.7.0 20110831 (experimental)

Compile and test:

$ g++ -Wall packed.cpp -o packed && ./packed
7
$ x86_64-w64-mingw32-g++ -Wall packed.cpp -o packed.exe
--> prints '8' when run on Windows.

The Linux binary prints the expected size of 7 bytes, the Windows binary 8 bytes. Why the difference?

My guess is the attribute somehow gets ignored (because of a bug). Use `offsetof(header, size)` to find out if it's so. — cnicutar, Oct 17 '11 at 05:26
If I compile it directly from windows xp with MinGW g++ 4.5.2, it prints 7. 32 bit though. — Vlad, Oct 17 '11 at 05:28
struct sizes varying is expected, which is why it is a good idea to never use/pass structures across compile domains. — old_timer, Oct 17 '11 at 05:28
If it's for an ID3 tag, I suppose the struct packing is part of the format, so no choice there. — Vlad, Oct 17 '11 at 05:31
@DrTwox: you're aware you're using an experimental compiler for the second part, right? — Mat, Oct 17 '11 at 05:47
Side note: Any C or C++ code that relies on specific struct member alignment (e.g. reading or writing structures directly to disk or network) smells bad... — 6502, Oct 17 '11 at 05:48
@Dani except that it is not universal and universally does not work. it is just a bad idea. 6502 obviously said it in a better way than I did. smells, bad, sounds bad, etc. — old_timer, Oct 17 '11 at 13:36

score 7 · Answer 1 · edited Jul 15 '16 at 18:20

7

gcc 4.7.0 does it this way to be compatible with 64-bit MSVC++. If you want to pack the structure properly, compile with -mno-ms-bitfields. (But then your layout will be incompatible with MSVC++.)

edited Jul 15 '16 at 18:20

Jonathan Wakely

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answered Oct 17 '11 at 08:15

TonyK

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score 4 · Accepted Answer · answered Oct 17 '11 at 05:33

4

Section 6.37.3 of the gcc attributes explains it as a difference in ABI specs, see here: http://gcc.gnu.org/onlinedocs/gcc/Type-Attributes.html

answered Oct 17 '11 at 05:33

Necrolis

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The microsoft compiler lets you pack the members as tight as you like. See: #pragma pack for the MS compiler. – harper Oct 17 '11 at 05:40

score 3 · Answer 3 · answered Oct 17 '11 at 05:49

The attribute((packed)) is compiler-specific to GCC. Hence, that code won't even compile with MSVC++. Maybe you used another compiler for Windows, though. However, with MSVC++ you could do this:

#include <stdint.h>
#include <cstdio>

// id3 header from an mp3 file
#pragma pack(push,1)
struct header
{    
        uint8_t version[ 2 ];
        uint8_t flags;
        uint32_t size;
};
#pragma pack(pop)

int main( int argc, char **argv )
{
        printf( "%u\n", (unsigned int)sizeof( header ) );
        return 0;
}

and the struct will be 7 bytes.

But it'll compile just fine using mingw under windows, which it appears the OP is using — Necrolis, Oct 17 '11 at 06:06

Jeegar Patel · Answer 4 · 2011-10-17T08:30:20.873

1

This is all about attribute and word alignment in memory

see if you write

struct header
{    
        uint8_t version[ 2 ];
        uint8_t flags;
        uint32_t size;
};

then linux & windows both have size 8

but when you specify attribute to avoid default world allignment then

struct header
{    
        uint8_t version[ 2 ];
        uint8_t flags;
        uint32_t size;
} __attribute__((packed));

then in linux because of attritube size becomes 7

see gcc spec says that

If packed is used on a structure, or if bit-fields are used 
it may be that the Microsoft ABI packs them differently than 
GCC would normally pack them.

edited Oct 17 '11 at 08:30

answered Oct 17 '11 at 05:41

Jeegar Patel

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Perhaps because the `compiler specific` bit does not matter ? What matters is the quote about ABI. – Matthieu M. Oct 17 '11 at 06:39
i have added that bcz user is going to compile program with attribute((packed)) in windows, attribute((packed)) is not supported in windows compiler – Jeegar Patel Oct 17 '11 at 06:48
2

"windows compiler"? OP specifies he uses GCC both on linux and Windows - MSVC isn't the only compiler around for Windows. – snemarch Oct 17 '11 at 08:21

score 0 · Answer 5 · answered Dec 20 '18 at 11:06

0

Update. Latest MinGW works fine.

Both g++ (i686-win32-dwarf-rev0, Built by MinGW-W64 project) 8.1.0 and g++ (x86_64-win32-seh-rev0, Built by MinGW-W64 project) 8.1.0 prints sizeof() of sample code is exactly equal 7 bytes.

answered Dec 20 '18 at 11:06

Alexey Esaulenko

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Why would the size of a packed structure be different on Linux and Windows when using gcc?

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