Is it possible to check if a user literal is defined for given type and argument?

Question

I want to check at compile-time if user literal _name is defined for type Ret and argument Arg. While I have half-solution, it require

Answer 1

With is_detected functions family, you may just do

template <typename T>
using has_literal_x_type = decltype(operator"" _x(std::declval<T>()));

template <typename Ret, typename T>
using has_literal_x = std::is_same<Ret, detected_t<has_literal_x_type, T>>;

And test it with

static_assert(!has_literal_x<one, char const*>::value, "unexpected");
static_assert(!has_literal_x<one, unsigned long long int>::value, "unexpected");
static_assert(!has_literal_x<two, char const*>::value, "unexpected");
static_assert(!has_literal_x<two, unsigned long long int>::value, "unexpected");
static_assert(!has_literal_x<int, char const*>::value, "unexpected");

Demo

Answer 2

Is it possible to check if an user literal is defined for given type and argument?

The (short) answer is yes.

---

As an example, you can use the following specialization in your example code:

template<class T, class S> 
struct has_literal_x <T, S,
      std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value>
    > : std::true_type
{ };

That quickly becomes:

#include <iostream>
#include <type_traits>
#include <utility>

struct one { };
struct two { };

//one operator"" _x(char const*) { return {}; }
//two operator"" _x(unsigned long long int) { return {}; }

template<class T, class S, class = void>
struct has_literal_x : std::false_type
{  };

template<class T, class S> 
struct has_literal_x <T, S, 
      std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value> 
    > : std::true_type
{ };

int main()
{  
    std::cout << has_literal_x<one, char const*>::value << std::endl;
    std::cout << has_literal_x<two, unsigned long long int>::value << std::endl;

    std::cout << has_literal_x<one, unsigned long long int>::value << std::endl;
    std::cout << has_literal_x<two, char const*>::value << std::endl;

    std::cout << has_literal_x<int, char const*>::value << std::endl;
}

The output is the expected one: 0 for all of them.

---

Another way to do that in C++14 (mostly inspired by this answer of @Jarod42) is by means of a template variable.
As an example:

template<typename T, typename S, typename = void>
constexpr bool has_literal_v = false;

template<typename T, typename S>
constexpr bool has_literal_v<T, S, std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value>> = true;

The main would become instead:

int main()
{  
    std::cout << has_literal_v<one, char const*> << std::endl;
    std::cout << has_literal_v<two, unsigned long long int> << std::endl;

    std::cout << has_literal_v<one, unsigned long long int> << std::endl;
    std::cout << has_literal_v<two, char const*> << std::endl;

    std::cout << has_literal_v<int, char const*> << std::endl;
}

I find it easy to read and that's a constexpr variable. What else?