using declaration of a specialized variadic template class

Question

Is it possible to define FloatType in such a way that I can declare a f1 as

FloatType f1;

instead of

FloatType<> f1;

If i try to use the former i get a

error: use of class template 'FloatType' requires template arguments

template <typename T, typename... Args>
class Type
{
};

template <typename... Args>
class Type<float, Args...>
{
};

template <typename... Args>
using FloatType = Type<float, Args...>;

int
main(int, char **)
{
    FloatType<> f1;
    FloatType<float> f2;

    return 0;
}

Answer 1

No, that is impossible. From the standard §14.3/4, emphasis mine:

When template argument packs or default template-arguments are used, a template-argument list can be empty. In that case the empty <> brackets shall still be used as the template-argument-list. [ Example:

  template <class T = char> class String;
  String<>* p; // OK: String<char>
  String* q;   // syntax error

  template <class ... Elements> class Tuple;
  Tuple<>* t; // OK: Elements is empty
  Tuple* u;   // syntax error

—end example ]

But then, what's wrong with writing FloatType<>? If seeing the empty brackes really irks you, you can introduce another alias for them, but that kind of obfuscates things:

using DefFloatType = FloatType<>;

Plus, it's more typing!