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Is it possible to create a template that infers the underlying enum class type of an enum class value template parameter? I am looking to do something like the following non-working pseudo-example:

enum class MyEnumClass { e1, e2, e3 };

template<enum class T X> MyTemplate { // wrong syntax. want to specify X, infer T
     using enum_class_type = T;
     static constexpr enum_class_type value = X;
};

Which I would like to be instantiatable with only one parameter, e.g. MyTemplate<MyEnumClass::e1>:

static_assert( std::is_same<MyTemplate<MyEnumClass::e1>::enum_class_type, MyEnumClass>::value );
static_assert( MyTemplate<MyEnumClass::e2>::value == MyEnumClass::e2 );

I know that the following works, but it requires me to explicitly specify the enum class type as a second template parameter:

template<typename EnumClass, EnumClass X>
struct EnumValue {
    using enum_class_type = EnumClass;
    static constexpr enum_class_type value = X;
};
// usage: EnumValue<MyEnumClass, MyEnumClass::e1>::enum_class_type
Ross Bencina
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1 Answers1

5

This isn't possible directly until c++17, where you can do

template <auto x>
class MyTemplate {
    using enum_type= decltype(x);
    static constexpr enum_type value = x;
};

Your use case is cited as the exact reason for the new feature. See here

Joseph Ireland
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