c++ - 嵌套类的嵌套类的访问权限

Question

在 C++ 中，嵌套类有权访问封闭类的所有成员。这是否也适用于嵌套类的嵌套类？

这段代码

#include <iostream>

class A
{
public:
    class B
    {
    public:
        B() { std::cout << A::x << std::endl; }

        class C
        {
        public:
            C() { std::cout << A::x << std::endl; }

        };

    };

private:
    static const int x { 0 };

};

int main()
{
    A::B b;

    A::B::C c;
}

在 g++ 7.2 上编译时没有警告。但是，我不清楚这是否符合标准。标准草案 (N4727 14.7) 说:

A nested class is a member and as such has the same access rights as any other member.

但是，在上面的例子中，C不是A的成员，它是成员的成员。这里的标准模棱两可吗？ g++ 行为是否可移植？

Answer 1

However, in the example above C is not a member of A, it is a member of a member.

是的，这是明确定义的行为；访问权限从 B 转移。

根据标准[class.access]/2 ,

A member of a class can also access all the names to which the class has access.

和[class.mem]/1 ,

Members of a class are data members, member functions, nested types, enumerators, and member templates and specializations thereof.

C是B的嵌套类，也是B的成员，那么C可以访问命名 B 可以访问的内容，包括 A::x。同理，C::C是C的成员，它可以访问C可以访问的名字，所以访问 C::C 中的 >A::x 没问题。