How to know underlying type of class enum?

Question

I have a variable declared as:

enum class FooEnum: uint64_t {}

and I would like to cast to its base-type, but I don\'t want to hardcode the

Answer 1

Here is another approach for when underlying_type is not present. This method doesn't attempt to detect the signed-ness of the enum, just give you a type of the same size, which is more than enough for a lot of situations.

template<int>
class TIntegerForSize
{
    typedef void type;
};

template<>
struct TIntegerForSize<1>
{
    typedef uint8_t type;
};

template<>
struct TIntegerForSize<2>
{
    typedef uint16_t type;
};

template<>
struct TIntegerForSize<4>
{
    typedef uint32_t type;
};

template<>
struct TIntegerForSize<8>
{
    typedef uint64_t type;
};

template<typename T>
struct TIntegerForEnum
{
    typedef typename TIntegerForSize<sizeof(T)>::type type;
};

Usage:

enum EFoo {Alpha, Beta};
EFoo f = Alpha;
TIntegerForEnum<EFoo>::type i = f;
TIntegerForEnum<decltype(f)>::type j = f;

Answer 2

Both Visual C++ 10.0 and MinGW g++ 4.6.1 lack std::underlying_type, but both accept this code:

template< class TpEnum >
struct UnderlyingType
{
    typedef typename conditional<
        TpEnum( -1 ) < TpEnum( 0 ),
        typename make_signed< TpEnum >::type,
        typename make_unsigned< TpEnum >::type
        >::type T;
};

Answer 3

Your guessed syntax is amazingly close. You're looking for std::underlying_type in <type_traits>:

#include <type_traits>
#include <cstdint>

enum class FooEnum: std::uint64_t {};

int main()
{
    FooEnum myEnum;
    uint64_t* intPointer = (std::underlying_type<FooEnum>::type*)&myEnum;
}

Answer 4

You can use this:

• std::underlying_type class template to know the underlying type of enum.

The doc says,

Defines a member typedef type of type that is the underlying type for the enumeration T.

So you should be able to do this:

#include <type_traits> //include this

FooEnum myEnum;
auto pointer = static_cast<std::underlying_type<FooEnum>::type*>(&myEnum);