c++ - 为什么这些具有外部链接的名称不表示同一实体？

Question

考虑以下代码片段:

#include <iostream>

int a;
void address_of_a(void) {
    std::cout << &a << std::endl;
}

namespace N {
    int a;
    void address_of_a(void) {
        std::cout << &a << std::endl;
    }   
};

int main(void) {
    address_of_a();
    N::address_of_a();
    return 0;
}

全局命名空间和命名空间 N 被赋予外部链接，因为 N4567 的 3.5 [basic.link] 第 4 段说 p>

An unnamed namespace or a namespace declared directly or indirectly within an unnamed namespace has internal linkage. All other namespaces have external linkage...

此外，::a和N::a也被赋予外部链接，因为它们不适用于3.5 [basic.link ] 第 3 段，

A name having namespace scope (3.3.6) has internal linkage if it is the name of

• a variable, function or function template that is explicitly declared static; or,
• a variable of non-volatile const-qualified type that is neither explicitly declared extern nor previously declared to have external linkage; or
• a data member of an anonymous union.

但对于 3.5 [basic.link] 第 4 段，

... A name having namespace scope that has not been given internal linkage above has the same linkage as the enclosing namespace if it is the name of

• a variable; or
• ...

这意味着它们继承了与全局命名空间和N相同的链接(=外部链接)。总之，它们应表示相同的实体，因为 3.5 [basic.link] 第 9 段说

Two names that are the same (Clause 3) and that are declared in different scopes shall denote the same variable, function, type, enumerator, template or namespace if

• both names have external linkage or both names have internal linkage and are declared in the same translation unit; and
• ...

但是，它们似乎表示不同的实体，因为它们的地址不一致。这是为什么？

Answer 1

请参阅第二个项目符号。

Two names that are the same (Clause 3) and that are declared in different scopes shall denote the same variable, function, type, enumerator, template or namespace if

• both names have external linkage or else both names have internal linkage and are declared in the same translation unit; and

• both names refer to members of the same namespace or to members, not by inheritance, of the same class; and

• when both names denote functions, the parameter-type-lists of the functions (8.3.5) are identical; and

• when both names denote function templates, the signatures (14.5.6.1) are the same.

::a 和 N::a 不引用相同命名空间的成员，因此它们不表示相同的变量。