Making functions: How to provide a simple interface when working with std::function

make_function.cpp

// See: http://kbokonseriousstuff.blogspot.fr/2013/04/making-functions-how-to-provide-simple.html?spref=tw
// From: http://coliru.stacked-crooked.com/view?id=665b6505b8cbe3631604d8300f9e6b27-50d9cfc8a1d350e7409e81e87c2653ba 

#include <iostream>
#include <functional>

// Only takes std::functions
template<typename R, typename... P>
void do_something(std::function<R(P...)> &func) {
    std::cout << &func << std::endl;
}

struct not_pointer_to_member_function{};

template<typename T>
struct pointer_to_member_function_traits {
    typedef not_pointer_to_member_function type;
};

template<typename C, typename R, typename... P>
struct pointer_to_member_function_traits<R (C::*)(P...) const> {
    typedef std::function<R(P...)> type;
};

struct not_callable{};

template<typename C>
struct callable_traits {
    typedef typename pointer_to_member_function_traits<decltype(&C::operator())>::type type;
};

template<typename R, typename... P>
struct callable_traits<R(P...)> {
    typedef std::function<R(P...)> type;
};

template<typename Func>
typename callable_traits<Func>::type
make_function(const Func &func) {
    return func;
}

void abc(){}

struct fo_abc {
    void operator()() const {}
};

int main () {
    std::function<void(int, int)> lamarck = [&](int a, int b){ std::cout << a << b; };

    auto func = make_function([](){}); // lambda
    auto func2 = make_function(abc); // free function
    auto func3 = make_function(fo_abc()); // function object
    do_something(func);
    do_something(func2);
    do_something(func3);
}