为什么 injected-class-name 有时不被视为 class 模板中的模板名称？

Question

In the following cases, the injected-class-name is treated as a template-name of the class template itself:

it is followed by <

it is used as a template argument that corresponds to a template template parameter

it is the final identifier in the elaborated class specifier of a friend class template declaration.

所以我尝试检查所有 3 个案例（另外在基本歧义的情况下，尽管我认为这在这里无关紧要）。

第一种情况看起来很简单。

问题是 - 为什么注释掉的示例不起作用？他们不支持 GCC 和 Clang，所以我不认为这是一个实现问题

template <template <class> class> struct A;

template <class T> struct Base {};

template <class T> struct Derived: Base<int>, Base<char>
{
    // #1
    typename Derived::Base<double> d;
    
    // #2
    
    // using a = A<Base>;

    using a = A<Derived::template Base>;

    // #3

    template<class U1>
    friend struct Base;

    // template<class U>
    // friend struct Derived::template Base;
};

上面的规则是否只适用于模板本身，而不适用于基础？如果是这样，碱基的规则是什么，尤其是最后两种情况？

Answer 1

这里的相关规则是[temp.local]/4:

A lookup that finds an injected-class-name ([class.member.lookup]) can result in an ambiguity in certain cases (for example, if it is found in more than one base class). If all of the injected-class-names that are found refer to specializations of the same class template, and if the name is used as a template-name, the reference refers to the class template itself and not a specialization thereof, and is not ambiguous. [ Example:
template <class T> struct Base { };
template <class T> struct Derived: Base<int>, Base<char> {
  typename Derived::Base b;             // error: ambiguous
  typename Derived::Base<double> d;     // OK
};
— end example ]

我认为这意味着这是一个 gcc 和一个 clang 错误。在：

using a = A<Base>;

找到的所有注入-class-名称确实指的是同一 class 模板 (Base<T>) 和该名称用作 模板名称 （因为它是模板模板参数的模板参数），所以这应该仅引用 class 模板本身而不是模糊的。

为什么 injected-class-name 有时不被视为 class 模板中的模板名称？

Why injected-class-name is sometimes not treated as a template name in a class template?

c++

templates

c++14

injected-class-name

c++17