unqualified-id 如何在函数调用中包含 unqualified-name？

Question

在 C++ 草案 ISO(N4901/2021) 6.5.4（参数相关名称查找）中我们有：

When the postfix-expression in a function call (7.6.1.3) is an unqualified-id, and unqualified lookup (6.5.3) for the name in the unqualified-id does not find any (1.1) — declaration of a class member, or (1.2) — function declaration inhabiting a block scope, or (1.3) — declaration not of a function or function template

我想不出一个 unqualified-id 包含 unqualified name 的例子（两者不同） . Draft给出了下面的例子，但是我说不出什么是(f)(s):

namespace N {
struct S { };
void f(S);
}
void g() {
N::S s;
f(s); // OK: calls N::f
(f)(s); // error: N::f not considered; parentheses prevent argument-dependent lookup
}

Answer 1

I can't figure out a example of an unqualified-id that contains a unqualified name(with the two being different).

措辞中没有说“不合格的名称”。并且似乎在 «name» 之前漏掉了 «component» 这个词。

这是一个例子：

namespace N
{
    struct S { };
    template<typename>
    void f(S);
}

void g()
{
    N::S s;
    f<int>(s); // OK: ADL finds N::f
}

一个unqualified-id这里是f<int>（这是一个template-id）并且执行查找它的（组件）名称 f.

[expr.prim.id.unqual]/2:

A component name of an unqualified-id U is
— U if it is a name or
— the component name of the template-id or type-name of U, if any.

[temp.names]/2

The component name of a simple-template-id, template-id, or template-name is the first name in it.

术语 «name» 在 [basic.pre]/4 中定义：

A name is an identifier ([lex.name]), operator-function-id ([over.oper]), literal-operator-id ([over.literal]), or conversion-function-id ([class.conv.fct]).

unqualified-id 如何在函数调用中包含 unqualified-name？

How can a unqualified-id contain a unqualified-name in a function call?

c++

function

language-lawyer

lookup