Is it possible to determine if a type is a scoped enumeration type?

Question

Is there a type trait, or is it possible to write a type trait is_scoped_enum<T> such that:

• if T is a scoped enumeration, is_scoped_enum<T>::value is true and
• if T is any other type, is_scoped_enum<T>::value is false

Answer 1

C++23 will be providing is_scoped_enum which can be used once it gets implemented. See this link for documentation: is_scoped_enum. I don't think clang supports this yet (see clang status for the latest info of what features are supported).

For now I am using a slightly simplified version of the answer above (using _v and _t) as well as an implementation of is_underlying:

// C++ 23 should include 'is_scoped_enum' and 'to_underlying' so the following
// code can be removed:

template<typename T, bool B = std::is_enum_v<T>>
struct is_scoped_enum : std::false_type {};

template<typename T>
struct is_scoped_enum<T, true>
  : std::integral_constant<
      bool, !std::is_convertible_v<T, std::underlying_type_t<T>>> {};

template<typename T>
inline constexpr bool is_scoped_enum_v = is_scoped_enum<T>::value;

template<typename T, std::enable_if_t<is_scoped_enum_v<T>, int> = 0>
[[nodiscard]] constexpr auto to_underlying(T x) noexcept {
  return static_cast<std::underlying_type_t<T>>(x);
}