infer lambda return type in template

Question

Hello i'm trying to create a function which will make a call to function provided as a parameter with some provided arguments and return its value, my current approach is as follows

  #include <functional>
  #include <iostream>
  #include <type_traits>


  template <typename Res,typename... T>
  auto func(typename std::common_type<std::function<Res(T...)>>::type f, T... values) -> decltype(f(values...)) {
    f(values...);
  }

  int fx(int x,int y){return x+y; }
  int main() {
>>  func([](int x, int y, int z) { std::cout << (x*y*z) << std::endl; }, 3, 6, 8);
    return 0;
  }

if i replace Res with void and remove the auto&&decltype it works, but somehow i'm not able to get the proper return type.

Answer 1

Best and simplest way is to accept any functor, not just std::function:

template <typename F,typename... T>
auto func(F f, T... values) -> decltype(f(values...)) {
    return f(values...);
}

Answer 2

Tossing in one extra alternative (as the above would be too much to digest sometimes).

With c++17 you can do:

template <typename F, typename... T, typename RES = std::invoke_result<F, T...>>
RES func(F f, T... values) {
    return f(values...);
}

and then there's also std::result_of for before C++17.

https://en.cppreference.com/w/cpp/types/result_of