C++11 type trait to differentiate between enum class and regular enum

Question

I'm writing a promotion template alias similar to boost::promote but for C++11. The purpose of this is to avoid warnings when retrieving arguments from varidic functions. e.g.

template <typename T>
std::vector<T> MakeArgVectorV(int aArgCount, va_list aArgList)
{
    std::vector<T> args;
    while (aArgCount > 0)
    {
        args.push_back(static_cast<T>(va_arg(aArgList, Promote<T>)));
        --aArgCount;
    }
    return args;
}

The Promote template alias promotes the type following the default argument promotion for variadic arguments: 1) An integer that's smaller than an int is promoted to int 2) A float is promoted to double

My problem is that a standard C++ enum can be promoted but a C++11 enum class is not promoted (compiler does not generate a warning). I want Promote to work with a regular enum but ignore a C++11 enum class.

How can I tell the difference between an enum class and an enum in my Promote template alias?

Philippeville

The real problem is that you're using va_args instead of a std::initializer_list and/or variadic templates. – Moores

Thanks for the tip but I have the va_list because I'm working with a C interface. – Philippeville

@Sam: Is my answer solving your problem? – Promptbook

@AndyProwl: Perfect answer, just what I was looking for! – Philippeville

Possible duplicate of Is it possible to determine if a type is a scoped enumeration type? – Annular

Answer 1

Here is a possible solution:

#include <type_traits>

template<typename E>
using is_scoped_enum = std::integral_constant<
    bool,
    std::is_enum<E>::value && !std::is_convertible<E, int>::value>;

The solution exploits a difference in behavior between scoped and unscoped enumerations specified in Paragraph 7.2/9 of the C++11 Standard:

The value of an enumerator or an object of an unscoped enumeration type is converted to an integer by integral promotion (4.5). [...] Note that this implicit enum to int conversion is not provided for a scoped enumeration. [...]

Here is a demonstration of how you would use it:

enum class E1 { };
enum E2 { };
struct X { };

int main()
{
    // Will not fire
    static_assert(is_scoped_enum<E1>::value, "Ouch!");

    // Will fire
    static_assert(is_scoped_enum<E2>::value, "Ouch!");

    // Will fire
    static_assert(is_scoped_enum<X>::value, "Ouch!");
}

And here is a live example.

ACKNOWLEDGEMENTS:

Thanks to Daniel Frey for pointing out that my previous approach would only work as long as there is no user-defined overload of operator +.

Promptbook

+1, but there is one cave-at: It only works as long as the author of some enum class E has not defined his own operator+(int,E). Fix it by adding a void dummy(int) and use decltype(dummy(std::declval<E>())). – Rockweed

@DanielFrey: Good point. Actually I could use a different operator, such as ^, to make it less likely to interfere with a user-defined operator overload – Promptbook

@AndyProwl: or call a function that takes an int. – Moores

Test the conversion to int as, as far as I know, it can not be provided by the user. – Rockweed

@DanielFrey: You're right, that makes the whole thing much simpler actually. – Promptbook

@Fanael: Good suggestion, thank you. Actually, it turns out std::is_convertible<> does the job, so that a simple alias template is enough to define the trait. – Promptbook

@AndyProwl: Now do one last edit and derive from std::integral_constant instead of this 90's static const bool value = ...-thingy. ;) – Rockweed

Answer 2

It appears that as of C++23 a similar solution to the one provided by @AndyProwl will be available from type_traits

#include <type_traits>

enum E { a, b };
enum class Es { x, y, z };

std::is_scoped_enum_v<E>;  // False
std::is_scoped_enum_v<Es>; // True

Riley