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?

Answer 1

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++.)

Answer 2

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

Answer 3

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.