C++ compile-time predicate to test if a callable object of type F can be called with an argument of type T

Question

I would like to create a compile-type function that, given any callable object f (function, lambda expression, function object, ...) and a type T,

Answer 1

A variant of Paul's answer, but following the standard SFINAE test pattern. Again a generic trait with arbitrary parameter types A...:

struct can_call_test
{
    template<typename F, typename... A>
    static decltype(std::declval<F>()(std::declval<A>()...), std::true_type())
    f(int);

    template<typename F, typename... A>
    static std::false_type
    f(...);
};

template<typename F, typename... A>
using can_call = decltype(can_call_test::f<F, A...>(0));

Then a constexpr function as you requested:

template<typename T, typename F>
constexpr bool is_callable_with(F&&) { return can_call<F, T>{}; }

Check live example.

This will work with functions, lambda expressions, or function objects with arbitrary number of arguments, but for (pointers to) member functions you'll have to use std::result_of<F(A...)>.

---

UPDATE

Below, can_call has the nice "function signature" syntax of std::result_of:

template<typename F, typename... A>
struct can_call : decltype(can_call_test::f<F, A...>(0)) { };

template<typename F, typename... A>
struct can_call <F(A...)> : can_call <F, A...> { };

to be used like this

template<typename... A, typename F>
constexpr can_call<F, A...>
is_callable_with(F&&) { return can_call<F(A...)>{}; }

where I've also made is_callable_with variadic (I can't see why it should be limited to one argument) and returning the same type as can_call instead of bool (thanks Yakk).

Again, live example here.

Answer 2

I would make a type trait first:

template<class X = void>
struct holder
{
    typedef void type;
};

template<class F, class T, class X = void>
struct is_callable_with_trait
: std::false_type
{};

template<class F, class T>
struct is_callable_with_trait<F, T, typename holder<
    decltype(std::declval<F>()(std::declval<T>()))
>::type>
: std::true_type
{};

And then if you want, you can turn it into a function:

template<typename T, typename F>
constexpr bool is_callable_with(F&&) 
{
    return is_callable_with_trait<F&&, T>::value;
}

Answer 3

template<class F, class T, class = void>
struct is_callable_with_impl : std::false_type {};

template<class F, class T>
struct is_callable_with_impl<F,T,
     typename std::conditional< 
              true,
              void,
              decltype( std::declval<F>() (std::declval<T>()) ) >::type
      > : std::true_type {};

template<class T, class F>
constexpr bool is_callable_with(F &&) 
{ 
     return is_callable_with_impl< F, T >::value; 
}

It is basically the same solution as the one posted by Paul, I just prefer to use conditional<true, void, decltype( ... ) > instead of an holder class to avoid namespace pollution.