c++ - ADL 未选取带有模板参数列表的后缀表达式

Question

我想了解为什么调用模板f下面的代码无法编译:

struct A
{
    template<class...>
    friend void f(A) { }
} x;

int main()
{
    f<>(x);
}

ADL 要求函数调用中的后缀表达式是非限定 ID。 simple-template-id ( f<> ) 不是一个 unqualified-id 吗？

Answer 1

相关条款似乎是7.3.1.2 [namespace.memdef]第3段:

Every name first declared in a namespace is a member of that namespace. If a friend declaration in a non-local class first declares a class, function, class template or function template the friend is a member of the innermost enclosing namespace. The friend declaration does not by itself make the name visible to unqualified lookup (3.4.1) or qualified lookup (3.4.3). ...

也就是说，找到该名称的唯一方法是通过 ADL。但是，要应用模板参数，需要根据 14.2 [temp.names] 第 2 段找到名称:

For a template-name to be explicitly qualified by the template arguments, the name must be known to refer to a template.