c++ functor and function templates

Question

consider this simple and pointless code.

#include <iostream>

struct A {
    template<int N>
    void test() {
        std::cout << N << std::endl;
    }
};

int main() {
    A a;
    a.test<1>();
}

It is a very simple example of a function template. What if however, I wanted to replace A::test with an overloaded operator() to make it a functor?

#include <iostream>

struct A {
    template<int N>
    void operator()() {
        std::cout << N << std::endl;
    }
};

int main() {
    A a;
    a<1>(); // <-- error, how do I do this?
}

Certainly if the operator() took parameters which were dependent on the template, the compiler could possibly deduce the template. But I just can't figure out the proper syntax to specify template parameters with a parameterless functor.

Is there a proper way to do this?

Obviously, this code would work since it bypasses the functor syntax:

a.operator()<1>();

but that kinda defeats the purpose of it being a functor :-P.

Answer 1

You can only call

a.operator()<1>();

but that would not be using a functor. Functors need a non template operator(), as they must be able to be called as varname() and that won't work with your code.

To make it a real functor change your code a template class (functors are classes):

#include <iostream>

template<int N>
struct A {
    void operator()() {
        std::cout << N << std::endl;
    }
};

int main() {
    A<1> a;
    a();
}

Answer 2

There's not another "direct" way I know other than the:

 a.operator()<1>();

syntax. If you're open to changing the code, moving the template parameter to the class would work, or using a (boost|tr1)::bind to make a (boost|tr1)::function object.