c++ - Derive from template constructor of template base class

Question

Just curious, is it ever possible to inherit from a template class and in constructor of the derived class, call constructor of the base class which is also templated and has no arguments to deduce its types from?

template<typename T>
struct Base {
    template<typename D>
    Base() {                // no argument of type D to infer from
        static_assert(std::is_same<T,D>::value, "");
    }
};

struct Derived : Base<int> {
    Derived()  : Base<int>::Base<int>() {} // is there a way to write it correctly?
};

I can replace template constructor by a template method in my particular case, but still it is an interesting question about the language flexibility.

Answer 1

What the C++ standard says about this (section 14.8.1):

[ Note: Because the explicit template argument list follows the function template name, and because conversion member function templates and constructor member function templates are called without using a function name, there is no way to provide an explicit template argument list for these function templates. — end note ]

It's a note, not a rule, because it actually is a consequence of two other rules, one in the same section:

Template arguments can be speci?ed when referring to a function template specialization by qualifying the function template name with the list of template-arguments in the same way as template-arguments are speci?ed in uses of a class template specialization.

and from 12.1

Constructors do not have names.