合格的朋友功能模板实例化 [英] Qualified friend function template instantiation

问题描述

查看 C ++模板：完整指南（第二版） ） / 官方网站今天早上，我遇到了一个部分，我不太明白（12.5.2，如果您有书）。忽略此处无关紧要的内容：

Looking at C++ Templates: The Complete Guide (2nd Edition)/official site this morning I came across a section I can't quite make sense out of (12.5.2 if you have the book). Omitting what seems irrelevant here:

如果[朋友声明中]的名称后没有尖括号，则有两种可能性

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

1. 如果名称不合格[...]

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

如果名称​​是合格的（它包含 :: ），则该名称必须引用先前声明的函数或函数模板。匹配功能优于匹配功能模板。但是，这样的朋友声明不能是定义。

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.

使用以下代码

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);
                               ^

c语错误：

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

我正试图探底为此，我已经重新键入了几次代码（不过如果有人拿着这本书，也可以再次输入）。规则正确，编译器错误吗？该代码不是该规则的正确证明吗？

I'm trying to get to the bottom of this, and I've retyped the code a few times (though again if someone has the book...). Is the rule right and the compilers are wrong? Is the code not the right demonstration of the rule?

完整的代码包括一个较早的朋友函数定义：

The complete code includes an earlier friend function definition:

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

使用clang接受和gcc拒绝（这是其他问题）。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。

Which clang accepts and gcc rejects (that's a different question). In either case it doesn't demonstrate the rule that the author states, it's the second one referencing the earlier one in the same class.

推荐答案

这本书是对的。 [temp.friend] p1 -突出显示相关部分：

The book is right. [temp.friend]p1 - with the relevant part highlighted:

对于不是模板声明的朋友函数声明：

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

• if朋友的名字是合格的或不合格的 template-id ，[...]

朋友的名字是 qualified-id ，并且在指定的类或命名空间中找到匹配的非模板函数[...]

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

如果朋友的名字是 qualified-id ，并且在指定的类或命名空间中找到了匹配的函数模板，则朋友声明将引用推导该功能模板的专业化

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

[...]

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