c++ - 合格的 friend 功能模板实例化

Question

今天早上看着C++ Templates: The Complete Guide (2nd Edition) / official site，我遇到了一个我不太明白的部分(如果您有这本书，则是12.5.2)。忽略这里无关紧要的内容:

If the name [in the friend declaration] is not followed by angle brackets there are two possibilities

1. If the name isn't qualified [...]

2. If the name is qualified (it contains ::), the name must refer to a previously declared function or function template. A matching function is preferred over a matching function template. However, such a friend declaration cannot be a definition.

用下面的代码

void multiply(void*);

template <typename T>
void multiply(T);

class Comrades {
    // ... skipping some friends that do not effect the error message
    friend void ::multiply(int); // refers to an instance of the template
    // ...
};

gcc错误:

error: ‘void multiply(int)’ should have been declared inside ‘::’
     friend void ::multiply(int);
                               ^

铛错误:

error: out-of-line declaration of 'multiply' does not match any
  declaration in the global namespace
friend void ::multiply(int);
              ^~~~~~~~

我试图弄清这件事的深处，并且我已经重新输入了几次代码(不过如果有人拿着这本书的话……)。规则正确，编译器错误吗？代码不是规则的正确证明吗？

完整的代码包括一个较早的 friend 函数定义:

class Comrades {
    friend void multiply(int) { }
    friend void ::multiply(int);
}

哪个clang接受而gcc拒绝(这是different question)。无论哪种情况，它都没有证明作者陈述的规则，这是第二个引用同一类中较早版本的规则。

Answer 1

这本书是对的。 [temp.friend]p1-突出显示相关部分:

For a friend function declaration that is not a template declaration:

• if the name of the friend is a qualified or unqualified template-id, [...]

• if the name of the friend is a qualified-id and a matching non-template function is found in the specified class or namespace, [...]

• if the name of the friend is a qualified-id and a matching function template is found in the specified class or namespace, the friend declaration refers to the deduced specialization of that function template

• [...]