Why is template name available in derived class (the base class is an instance of the template)?

Question

I came across this code (simplified with basic types):

template <typename T>
class Base {
  T t;
};

class Derived : public Base<short> {
 public:
  using Base<short>::Base;
};

int main() {
  Derived::Base<long long> x;
  printf("%lu\n", sizeof(x));
  return 0;
}

It compiles and works (output is 8, which is the size of long long). It seems I can get Base<T> for any type T using Derived::Base, even if Derived is just a subclass of Base<short>. (In the code I came across, Base itself is not visible to main.)

However, I don't quite understand this grammar and why it works.

Is Derived::Base a template name, or a class, or a function (ctor)? It seems like a template name. Is the template name available in all classes that instantiate this template (like template name Base is in Base<T> for all type T)? I'm so confused. Any explanation or pointer to cppreference or the C++ standard are appreciated.

Answer 1

From en.cppreference.com/injected-class-name:

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

• it is followed by <
• [..]

So Base inside Base<T> is, depending of context, a class or a template name.