为什么在使用gcc时在Linux和Windows上打包结构的大小会不同？ [英] Why would the size of a packed structure be different on Linux and Windows when using gcc?

问题描述

在下面的代码中，为什么在使用gcc编译时在Linux和Windows上打包结构的大小不同？

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;
}

使用的gcc版本：

$ 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)

编译和测试：

$ 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.

Linux二进制打印预期的7字节大小，Windows二进制8字节。为什么有区别？

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

推荐答案

第6.37.3节的gcc属性将它解释为ABI规范的区别， a href =http://gcc.gnu.org/onlinedocs/gcc/Type-Attributes.html =nofollow> http://gcc.gnu.org/onlinedocs/gcc/Type-Attributes.html

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

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