c++ - 标准布局类上的偏移量？

Question

#include <iostream>
#include <cstddef>
class Foo
{
    int a;
    int b;
    float c;
};
int main()
{
    Foo foo;
    std::cout << offsetof(Foo, b) << std::endl;
    return 0;
}

以上代码无法使用 gcc-4.8.2 或 vc++11 编译。错误消息是无法访问类 Foo 中的私有(private)成员 b。

但是按照标准，offsetof应该支持标准布局类，而Foo是标准布局类。

这是gcc-4.8.2或vc++11的缺陷，还是我对c++标准的理解有误？

Answer 1

offsetof 被定义为一个宏，因此它无法绕过访问控制并获得私有(private)成员的访问权限，我们可以看到这就是起草 C++ 标准部分 17.6.1.2 Headers 段落 5 的情况，其中说(强调我的 ):

Names which are defined as macros in C shall be defined as macros in the C++ standard library, even if C grants license for implementation as functions. [ Note: The names defined as macros in C include the following: assert, offsetof, setjmp, va_arg, va_end, and va_start. —end note ]

更新

所以有一些 hack 可以让您以标准方式访问类的私有(private)成员，但是如果我们回到 C99 草案标准，C++ 标准草案又回到了 offsetof然后我们在 7.17 Common definitions 段 3 中看到(强调我的):

   offsetof(type, member-designator)

which expands to an integer constant expression that has type size_t, the value of which is the offset in bytes, to the structure member (designated by member-designator), from the beginning of its structure (designated by type). The type and member designator shall be such that given

   static type t;

then the expression &(t.member-designator) evaluates to an address constant.

如果您试图从类外访问一个私有(private)成员，情况就不会如此。