Achieve functor overloading through composition - c++

Question

Given some existing functors:
struct incr {
int operator()(int x) const { return x + 1; }
};
struct rep_str {
std::string operator()(const std::string& s) const { return s + s; }
};
I'm wondering if it's possible to achieve something like this:
auto f = overload<incr, rep_str>();
f(1); // returns 2
f("hello"); // returns "hellohello"
Multiple overloads may look like:
auto f = overload<fa, fb, fc, ...>();
// or...
auto g = overload<fa, overload<fb, overload<fc, ...>>>();
I'm thinking maybe use SFINAE with std::result_of_t or something like that, but haven't figured out how.

Answer 1

You don't need anything too fancy: just inherit from all the arguments and use using-declarations to bring in operator() from the base classes. However, in the variadic case, you can't have a pack expansion in a using-declaration, so you have to use a recursive approach, like so:
template <class... Ts>
struct overload {}; // only used for empty pack
template <class T>
struct overload<T> : private T {
using T::operator();
};
template <class T1, class T2, class... Ts>
struct overload<T1, T2, Ts...> : private T1, overload<T2, Ts...> {
using T1::operator();
using overload<T2, Ts...>::operator();
};

Answer 2

Brian's answer is better, IMHO, but since I worked on it, here's mine:
#include <type_traits>
#include <utility>
template <typename... Fns>
struct overload;
template <typename Fn, typename... Fns>
struct overload<Fn, Fns...>
{
template <typename... T>
std::result_of_t<Fn(T...)> operator()(T && ... args) const {
return Fn()(std::forward<T>(args)...);
}
using next = overload<Fns...>;
template <typename... T>
std::result_of_t<next(T...)> operator()(T && ... args) const {
return next()(std::forward<T>(args)...);
}
};

Answer 3

this can be done using template specialization:
#include <string>
#include <iostream>
template <typename...Args>
struct overload{
};
template <> struct overload<int>{
int operator()(int x) const { return x + 1; }
};
template <> struct overload< std::string>{
std::string operator()(const std::string& s) const { return s + s; }
};
template <typename...Args >
auto f(Args...arg){
overload<Args...> func;
return func(arg...);
}
int main()
{
std::cout << f(3) << std::endl << f(std::string("Hello"));
}
Note: two answers by #Brian and #md5i more general and elegant and perfect and better than this.