c++ - 类模板中可见的友元函数名称

Question

考虑以下示例:

template <typename T>
class C
{
    public:
        friend void f() {}
        friend void f(C<T>) {}
};

C<int> c;

void g(C<int>* p)
{
     f(); 
     f(*p);
}

使用 GCC 5.2 编译会引发以下编译错误:

no matching function for call to 'f()'

但是标准在 14.6.5 中说:

Friend classes or functions can be declared within a class template. When a template is instantiated, the names of its friends are treated as if the specialization had been explicitly declared at its point of instantiation.

为什么编译失败？在 GCC 3.4 中，它通过了。

Answer 1

f只能通过参数相关名称查找 (ADL) 找到。第二个调用编译是因为 p 的指针对象作为参数传递的 的类型为 C<int> - 这会导致 ADL 介入并检查其他不可见的功能。事实上，第一次重载f根本找不到，因为无法将关联传达给 C 的任何特化.

只需查看您后面的报价，[temp.inject]/2 :

As with non-template classes, the names of namespace-scope friend functions of a class template specialization are not visible during an ordinary lookup unless explicitly declared at namespace scope (11.3). Such names may be found under the rules for associated classes (3.4.2). ¹⁴¹

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¹⁴¹⁾ Friend declarations do not introduce new names into any scope, either when the template is declared or when it is instantiated.