C++类模板函数可以访问嵌套类私有(private)成员

Question

以下代码无法按预期编译:

#include<iostream>
class Enclosing {
    int x;
    class Nested { int y; };

    void EnclosingFun(Nested *n) {
        std::cout << n->y;  // Compiler Error: y is private in Nested
    }
};

但是，如果我将 EnclosingFun 更改为模板成员函数，编译器 (gcc-7) 不会提示访问 y:

#include<iostream>
class Enclosing {      
    int x;
    class Nested { int y; };

    template <typename T>
    void EnclosingFun(Nested *n, T t) {
        std::cout << t << n->y;  // OK? Why?
    }      
};

这是 gcc 中的错误吗？还是c++对模板成员函数访问嵌套类有不同的访问规则？

Answer 1

Is this a bug in gcc? Or does c++ have different access rules for template member function to access nested classes?

也没有。

根据标准，§17.8.1 Implicit instantiation [temp.inst] ,

An implementation shall not implicitly instantiate a function template, a variable template, a member template, a non-virtual member function, a member class, a static data member of a class template, or a substatement of a constexpr if statement ([stmt.if]), unless such instantiation is required.

这意味着，Enclosing::EnclosingFun() 未在此处实例化。将调用添加到它会导致它被实例化，然后你会得到错误，例如

prog.cc:8:30: error: 'int Enclosing::Nested::y' is private within this context
         std::cout << t << n->y;  // OK? Why?
                           ~~~^

LIVE