How to generalize a template specialization for any variadic template type?

Question

I have the following type function to calculate whether some type T is part of the list of types in an std::tuple:

template<typename T, typename Tuple>
struct IsInTuple;

template<typename T, typename ...Ts>
struct IsInTuple<T, std::tuple<Ts...>>
{
  static constexpr bool value = std::disjunction_v<std::is_same<T, Ts>...>;
};

My question is, is it possible to generalize this function for any variadic template type taking a variadic list of types so that it not only works for std::tuple's, but for example also for std::variant's?

Answer 1

Is it possible to generalize this function for any variadic template type taking a variadic list of types so that it not only works for std::tuples, but for example also for std::variants?

Yes, you can. Just use template template parameter to generalize the type traits.

template<typename T, typename Class> struct IsInTypeList;

template<typename T
       , template<typename...> class Class, typename... Ts>
struct IsInTypeList<T, Class<Ts...>>
{
    static constexpr bool value = (std::is_same_v<T, Ts> || ...);
                            // or std::disjunction_v<std::is_same<T, Ts>...>;
};

// Example usage
static_assert(IsInTypeList<int, std::tuple<int, float, double>>::value, "int is not in the tuple");
static_assert(!IsInTypeList<char, std::tuple<int, float, double>>::value, "char is in the tuple");
static_assert(!IsInTypeList<char, std::variant<int, float, double>>::value, "char is in the variant");

See live demo