Disable branch using "if constexpr" and SFINAE

Question

I want to enable / disable branches at compile time depending on whether it is possible to call a function with specific arguments. What needs to be done in a condition if constexpr? I can get the result type through std::result_of(decltype(add)(A, B)), but how can I check if the result type is really? (i.e. how to convert this information to bool?)

const auto add = [](const auto a, const auto b) { return a + b; };

const auto subtract = [](const auto a, const auto b) { return a - b; };

template <typename A, typename B>
void foo(A a, B b) {

  if constexpr ( /* can add(a, b) be called? */ ) {
    std::cout << "result of add: " << add(a, b) << std::endl;
  }

  if constexpr ( /* can subtract(a, b) be called? */ ) {
    std::cout << "result of subtract: " << subtract(a, b) << std::endl;
  }
}

Answer 1

First you need to make your SFINAE-friendly landbat.

#define RETURNS(...)\
  noexcept(noexcept(__VA_ARGS__))\
  ->decltype(__VA_ARGS__)\
  { return __VA_ARGS__; }

const auto add = [](const auto a, const auto b) RETURNS( a + b );

const auto subtract = [](const auto a, const auto b) RETURNS( a - b );

Now adding and peeling can be tested in the context of SFINAE.

namespace details {
  template<class, class, class...>
  struct can_invoke:std::false_type {};
  template<class F, class...Args>
  struct can_invoke<F, std::void_t< std::result_of_t< F&&(Args&&...) > >, Args... >:
    std::true_type
  {};
}
template<class F, class...Args>
using can_invoke_t = details::can_invoke<F, Args...>;

template<class F, class...Args>
constexpr can_invoke_t< F, Args... >
can_invoke( F&&, Args&&... ){ return {}; }

and we are ready:

template <typename A, typename B>
void foo(A a, B b) {

  if constexpr ( can_invoke( add, a, b ) ) {
    std::cout << "result of add: " << add(a, b) << std::endl;
  }

  if constexpr ( can_invoke( subtract, a, b ) {
    std::cout << "result of subtract: " << subtract(a, b) << std::endl;
  }
}

this C ++ 14; atC ++ 11 it's more uncomfortable in C ++ 17 , ( , , add std::invoke).

---

++ 17 :

template<class F>
constexpr auto invoke_test( F&& ) {
  return [](auto&&...args) ->
  can_invoke_t<F, decltype(args)...>
  { return {}; };
}

template <typename A, typename B>
void foo(A a, B b) {

  if constexpr ( invoke_test( add )( a, b ) ) {
    std::cout << "result of add: " << add(a, b) << std::endl;
  }

  if constexpr ( invoke_test( subtract )( a, b ) {
    std::cout << "result of subtract: " << subtract(a, b) << std::endl;
  }
}

invoke_test , " , ".

Answer 2

SFINAE , . can_be_called :

#include <type_traits>

template<class Func, class... Args>
constexpr auto
can_be_called(Func&& func, Args&&... args)
    -> decltype(
            (std::forward<Func>(func)(std::forward<Args>(args)...)
            , bool{}))
{ return true; }

struct Dummy {
    template<class T> constexpr Dummy(T&&) {}
};

template<class... Args>
constexpr bool 
can_be_called(Dummy, Args&&...) { return false; }

// test
#include <iostream>
void foo(int, int) {}

struct A{};

int main() {
    if constexpr( can_be_called(foo, 1, 2) ) {
        std::cout << "OK\n";
    }

    if constexpr ( !can_be_called(foo, A{}, 2) ) {
        std::cout << "NO\n";
    }
}