Member function & const member function pointer deduction

Question

I have the following code:

template <class Ret>
class Foo
{
public:
    template <class T>
    void foo(T&, const std::function<Ret()>&)
    {
        std::cout << "std::function<Ret()>\n";
    }
    template <class T>
    void foo(T&, Ret(T::*)() const)
    {
        std::cout << "Ret(T::*)() const\n";
    }
    template <class T>
    void foo(T&, Ret(T::*)())
    {
        std::cout << " Ret(T::*)()\n";
    }
};

class A
{
public:
    void foo1() const
    {
    }
    void foo()
    {
    }
};

 int main()
 {

     A a;
     Foo<void> f;
     f.foo(a, &A::foo);
     f.foo(a, &A::foo1);
     f.foo(a, std::bind(&A::foo, a));
 }

It works well, but I don't want to have 2 different function for const and non-const member function pointer. So the question is: is there a way to merge void foo(T&, const std::function<Ret()>&) and void foo(T&, Ret(T::*)() const) to one function? Note, that there is std::function overload which should also participate in resolution after merge. I need some function which will take member function pointers only. And all other will make its way to std::function version.

Answer 1

You seem to be asking one implicit question and one explicit question. The answer to your implicit question, "how to merge the const and nonconst versions", is as follows

template<typename T, typename U, typename enable_if<is_fuction<T>::value, int>::type = 0> 
void takesboth(T U::*);