Use invoke_result with void argument type?

Question

I'm trying to do the following:

struct Unwrapper
{
    template<typename T>
    auto operator()(const T& arg, std::enable_if_t<isPrimitive<T>, void>* = nullptr) {return arg;}

    template<typename T>
    auto operator()(const T& arg, std::enable_if_t<!isPrimitive<T>, void>* = nullptr) {return arg.wrapped();}

    void operator()(void) {}
};

template<typename T>
using UnwrappedT = std::invoke_result_t<Unwrapper, T>; // error: no type named ‘type’ in ‘struct std::invoke_result<Unwrapper, void>’

The docs for std::invoke_result suggests it should work for Args being void (i.e. none), specifically it says the void case not working was a "quirk" of the now deprecated std::result_of.

But no, void doesn't work. It kind of makes sense because one also can't do std::declval<T>() for T = void, and std::invoke_result is supposed to be implemented in terms of std::declval.

Question is, what's the most elegant/direct way to patch the code to work with void? I could do something with std::conditional but I expected better. (using C++17)

Relevant question: this, this.

Answer 1

You could do this:

template<typename... T>
using UnwrappedT = std::invoke_result_t<Unwrapper, T...>; 

UnwrappedT<> would handle the void case.

If you want UnwrappedT<void> to mean UnwrappedT<>, you'll need some way of dropping the void. conditional is the most familiar way of doing thay:

template<typename T>
using UnwrappedT = typename std::conditional_t<
    std::is_void_v<T>,
    std::invoke_result<Unwrapper>,
    std::invoke_result<Unwrapper, T>>::type;

Or you could have some fun with Boost.Mp11:

template<typename T>
using UnwrappedT = mp_apply<std::invoke_result_t,
    mp_remove_if<mp_list<Unwrapper, T>, std::is_void>>;

Answer 2

Have you tried ?

template<typename T>
using UnwrappedT = std::result_of_t<Unwrapper(const T&)>;