匿名结构中声明的函数的名称解析

Question

f 定义的参数列表中的以下代码中的 A、B、C 的名称解析应该起作用吗？

namespace ns
{
struct A {};
struct S
{
    struct B {};
    struct
    {
        struct C {};
        void f(A, B, C);
    } x;
};
}

#include <type_traits>

void std::type_identity_t<decltype(ns::S::x)>::f(A, B, C) {}

int main()
{
}

实际上 works 最近 clang。

Answer 1

[basic.lookup.unqual]/8 For the members of a class X, a name used ... in the definition of a class member outside of the definition of X, following the member's declarator-id(24), shall be declared in one of the following ways:
...
(8.2) — shall be a member of class X ..., or
(8.3) — if X is a nested class of class Y (11.4.10), shall be a member of Y,... or
...
(8.5) — if X is a member of namespace N, or is a nested class of a class that is a member of N, ... before the use of the name, in namespace N or in one of N's enclosing namespaces.

Footnote 24) That is, an unqualified name that occurs, for instance, in a type in the parameter-declaration-clause or in the noexcept-specifier.

这里X是decltype(ns::S::x)，Y是S，N是ns。因此，通过 (8.5) 找到 A，通过 (8.3) 找到 B，通过 (8.2) 找到 C。