class 模板的朋友 function 是否應該成為所有實例化的朋友？

Question

考慮以下代碼段：

template<typename T> struct Foo {
    friend void bar(Foo, Foo<char> f) {
        static_cast<void>(f.private_field);  // Should only compile when friends with Foo<char>.
    }
private:
    int private_field{};
};

int main() {
    bar(Foo<char>{}, Foo<char>{});  // Compiles.
    bar(Foo<bool>{}, Foo<char>{});  // Compiles erroneously?
}

它可以使用主干（截至 2020 年 5 月 6 日）GCC 成功編譯，但不能使用 Clang 和 MSVC 編譯：請參閱Godbolt 。

誰在這里？

來自 Clang 和 MSVC 的錯誤消息與預期一致：

<source>:3:29: error: 'private_field' is a private member of 'Foo<char>'

        static_cast<void>(f.private_field);  // Should only compile when friends with Foo<char>.

                            ^

<source>:11:5: note: in instantiation of member function 'bar' requested here

    bar(Foo<bool>{}, Foo<char>{});  // Compiles erroneously?

    ^

<source>:6:9: note: declared private here

    int private_field{};

        ^

1 error generated.

Compiler returned: 1

和

example.cpp

<source>(3): error C2248: 'Foo<char>::private_field': cannot access private member declared in class 'Foo<char>'

<source>(6): note: see declaration of 'Foo<char>::private_field'

<source>(2): note: see declaration of 'Foo<char>'

<source>(2): note: while compiling class template member function 'void bar(Foo<bool>,Foo<char>)'

<source>(11): note: see reference to function template instantiation 'void bar(Foo<bool>,Foo<char>)' being compiled

<source>(11): note: see reference to class template instantiation 'Foo<bool>' being compiled

Compiler returned: 2

Answer 1

GCC 在這里顯然是錯誤的：因為每個專業化定義了自己單獨的朋友 function （必須避免重復定義錯誤）， bar(Foo<char>,Foo<char>)是Foo<char> <char> 的朋友，但bar(Foo<char>,Foo<char>)不是。 該標准包含一個相關示例，其中提到了一個非常相似的案例。