Avoid angle brackets in default template

问题

If I have a template class with a default template type, I have to write the template angle brackets. Is it somehow possible to avoid this?

Example:

template <typename T=int>
class tt {
public:
  T get() { return 5; }
};

...

tt<> t;  // how to avoid <>
std::cout << t.get() << std::endl;

Until now i've did this by a separate namespace and redeclaring the class:

namespace detail_ {
template <typename T=int>
class tt {
public:
  T get() { return 5; }
};
}

class tt : public detail_::tt {}

...

tt t;
std::cout << t.get() << std::endl;

The problem is, if I want to use the class with an other type I have to go over namespace detail_. Is there another solution, which I didn't see yet.

回答1:

Since C++17, because of class template argument deduction, things have changed.

tt and tt<> are not the same thing: types and class templates were different and continue to be treated differently.

Anyway in simple scenarios like the one in your example, C++17 assumes what you mean and the <> aren't needed anymore.

Further details:

Template default arguments (specifically https://stackoverflow.com/a/50970942/3235496);
Why is <> required when specifying a template class which has defaults for all its template parameters?

回答2:

... if I want to use the class ...

This is a common source of confusion. A class template is not a class, but a template from which classes are generated. The angle brackets is what tells the compiler that you want to generate a class out of the class template with the given template arguments, without the angle brackets what you have is a template.

template <typename T = int>
struct TemplateClass {...};

template <template class T<typename> >
void f() {
   T<int> t; ...
}
template <typename T>
void g() {
   T t; ...
}

f<TemplateClass>();     // Accepts a template with a single type argument
g<TemplateClass<> >();  // Accepts a type, that can be generated out of the template

The language does not allow the coexistence of a template and a type with the same name in the same namespace, so the answer is that it cannot be done. You can create a type alias but you will have to give it a different name.