朋友功能在班级中不可见 [英] Friend function is not visible in the class

问题描述

我有以下代码：

struct M {
    friend void f() {}
    M() {
        f(); // error: 'f' was not declared in this scope
    }
};

int main() {
    M m;
}

实时示例

g ++ 4.8和clang3.4均无法编译，因为 f 在 M 内部不可见，或者这样说。

Both g++4.8 and clang3.4 fail to compile it, because f is not visible inside M, or so they say.

但是，该标准给出了类似代码的示例

However, the Standard gives an example of a similar code

class M {
  friend void f() { } // definition of global f, a friend of M,
                      // not the definition of a member function
};

并说

在一个类中定义的 friend 函数在定义它的
类的（词法）范围内。

A friend function defined in a class is in the (lexical) scope of the class in which it is defined.

（ISO / IEC 14882：2011 11.3 Friends [class.friend] p6，p7）

(ISO/IEC 14882:2011 11.3 Friends [class.friend] p6, p7)

由此我无法理解编译器如何找不到 f ，它是在使用它的同一类中定义的。

From this I can't understand how compiler can't find f which is defined in same class where it's used.

这两个编译器都有相同的错误是不太可能的。

那么，我错过了什么？

It's kinda unlikely that both compilers have the same bug.
So, what did I miss?

推荐答案

friend声明指出，周围名称空间中的名为 f 的函数是该类的朋友；但不会在名称空间中引入名称 f 。除非在命名空间中声明了它，否则它不可用（除非依赖于参数的查找）。

The friend declaration states that a function called f in the surrounding namespace is a friend of the class; but it does not introduce the name f into the namespace. It's not available (except by argument-dependent lookup) until it's been declared in the namespace.

相关规则是C ++ 11 7.3.1.2/3：

The relevant rule is C++11 7.3.1.2/3:

如果非本地类中的 friend 声明首先声明了一个类或函数，朋友类或函数是最内部的封闭名称空间的成员。 通过无限定查找或限定查找无法找到朋友的名字，直到在该命名空间范围内提供匹配的声明为止。

If a friend declaration in a non-local class first declares a class or function the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup or by qualified lookup until a matching declaration is provided in that namespace scope.

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