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

@Nawaz: A scoped enum is what an enum class is called in C++11, yes. — James McNellis
– James McNellis, Commented May 23, 2012 at 17:18
Just out of curiosity, what practical applications are there for this one? — Xeo
– Xeo, Commented May 24, 2012 at 18:43
@Xeo: I am overloading the bitwise operators for a set of scoped enumeration types. — James McNellis
– James McNellis, Commented May 24, 2012 at 21:08
bitbucket.org/martinhofernandes/wheels/src/353fc67489dc/include/… — R. Martinho Fernandes
– R. Martinho Fernandes, Commented May 24, 2012 at 21:15

R. Martinho Fernandes · Accepted Answer · 2012-05-23 18:05:34Z

38

I think testing if it is an enum and not implicitly convertible to the underlying type should do the trick.

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

template <typename T>
struct is_scoped_enum<T, true>
: std::integral_constant<bool,
    !std::is_convertible<T, typename std::underlying_type<T>::type>::value> {};

edited May 23, 2012 at 18:05

answered May 23, 2012 at 17:18

R. Martinho Fernandes

236k73 gold badges443 silver badges518 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

kennytm Over a year ago

Better use std::underlying_type<T> instead of int. An enum class in C++11 can base on something not convertible to int.

James McNellis Over a year ago

@KennyTM: What type? C++11 §7.2/2 states "The enum-base shall name an integral type;" is there an integral type not convertible to int?

kennytm Over a year ago

@JamesMcNellis: You're correct. Sorry for confusion. (I was thinking of is_convertible doing implicit conversion.)

James McNellis Over a year ago

This is a good solution, except that with the latest edit, the logic is now backwards: ::value is true if T is an unscoped enumeration. :-)

underscore_d Over a year ago

@Excelcius It's worse than the possibility of getting an error or not: using std::underlying_type with a non-enum type is undefined behaviour, which requires no diagnostic, so the compiler need not report the misuse, which then gives your program UB. I didn't look at common implementations but presume the Standard allows UB to facilitate easier implementation in valid use cases.

|

Dan H · Accepted Answer · 2022-02-12 14:28:01Z

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);
}

Collectives™ on Stack Overflow

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

2 Answers 2

7 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

7 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related