c++ - 如何在 vs 和 gcc 中使用 pack

Question

直到今天，我使用以下技术来打包结构:

#pragma pack(push,1)
struct mystruct
{
     char a1;
     char a2;
     int a3;
}
#pragma pack(pop)

mystruct mydata()
{ 
   mystruct ms;
   ms.a1='a';
   ms.a2='b';
   ms.a3=12;
   return ms;
 }

并假设ms打包为1，但今天有人告诉我，在上面的定义中，ms被打包为4，因为pack对定义没有影响，但对声明有影响。 http://msdn.microsoft.com/en-us/library/aa273913%28v=vs.60%29.aspx

谁能澄清我做的是否正确？

Answer 1

标准规定了§3.1/2

A declaration is a definition unless it declares a function without specifying the function’s body (8.4), it contains the extern specifier (7.1.1) or a linkage-specification25 (7.5) and neither an initializer nor a functionbody, it declares a static data member in a class definition (9.2, 9.4), it is a class name declaration (9.1), it is an opaque-enum-declaration (7.2), it is a template-parameter (14.1), it is a parameter-declaration (8.3.5) in a function declarator that is not the declarator of a function-definition, or it is a typedef declaration (7.1.3), an alias-declaration (7.1.3), a using-declaration (7.3.3), a static_assert-declaration (Clause 7), an attributedeclaration (Clause 7), an empty-declaration (Clause 7), or a using-directive (7.3.4).

因此你有一个你正确指出的结构定义，但它也是一个声明

pack has no effect on definitions

适用，但不适用于声明。事实上，MSVC 和 gcc/clang 正确地用 1 打包了上面的内容

struct mystruct_not_packed
{
    char a1;
    char a2;
    int a3;
};

#pragma pack(push,1)
struct mystruct
{
    char a1;
    char a2;
    int a3;
};

mystruct_not_packed object; // This doesn't apply
#pragma pack(pop)

int main(int argc, char *argv[])
{
    std::cout << sizeof(mystruct) << std::endl; // 6
    std::cout << sizeof(mystruct_not_packed) << std::endl; // 8
    std::cout << sizeof(object) << std::endl; // 8
}

(使用 MSVC2013U4 测试)

Example with gcc