Is it possible to supply template parameters when calling operator()?

Question

I'd like to use a template operator() but am not sure if it's possible. Here is a simple test case that won't compile. Is there something wrong with my syntax, or is this simply not possible?

struct A {
  template<typename T> void f() { }
  template<typename T> void operator()() { }
};

int main() {
  A a;
  a.f<int>();           // This compiles.
  a.operator()<int>();  // This compiles.
  a<int>();             // This won't compile.
  return 0;
}

Answer 1

Like chris mentioned in the comments, no, not with the shorthand syntax. You must use the full .operator()<T>() syntax;

Answer 2

If you really want to use templated operator() and want to avoid writing constructions like a.operator()<int>(); you can add an auxiliary parameter to it:

template <typename T>
struct type{};

struct A
{
    template<typename T>
    void operator()(type<T>) { }
};

int main()
{
    A a;

    a(type<int>());
}

Live demo.

---

In C++14 you can even omit empty brackets in a(type<int>()); via a variable template:

template <typename T>
struct type_{};

template <typename T>
constexpr type_<T> type{};

struct A
{
    template<typename T>
    void operator()(type_<T>) { }
};

int main()
{
    A a;

    a(type<int>);
}

Live demo.