Can an enum class be converted to the underlying type?

116
votes

Is there a way to convert an enum class field to the underlying type? I thought this would be automatic, but apparently not.

enum class my_fields : unsigned { field = 1 };

unsigned a = my_fields::field;

That assignment is being rejected by GCC. error: cannot convert 'my_fields' to 'unsigned int' in assignment.

4
c++c++11
If you want to convert to underlying type then use enum. – Pubby
FYI, this rule is defined in [C++11: 7.2/9]. – Lightness Races in Orbit
@Pubby Sadly unscoped 'enum' pollutes the outer scope with all the enumerants. Alas there is no best of both worlds (as of C++14 anyway) which cleanly nests scope while also implicitly converting to the base type (which is rather inconsistent with how C++ handles other class inheritance, when you pass a more derived type by value or reference to a function taking a base type). – Dwayne Robinson
@DwayneRobinson Yes there is. Stick an unscoped enum inside a struct or (more preferably) a namespace. Thus it is scoped and still has the implicit int conversion. (Though I'd be sure to think twice about why you need to convert to an int and perhaps consider if there's a better approach.) – Pharap

4 Answers

188
votes

I think you can use std::underlying_type to know the underlying type, and then use cast:

#include <type_traits> //for std::underlying_type

typedef std::underlying_type<my_fields>::type utype;

utype a = static_cast<utype>(my_fields::field);

With this, you don't have to assume the underlying type, or you don't have to mention it in the definition of the enum class like enum class my_fields : int { .... } or so.

You can even write a generic convert function that should be able to convert any enum class to its underlying integral type:

template<typename E>
constexpr auto to_integral(E e) -> typename std::underlying_type<E>::type 
{
   return static_cast<typename std::underlying_type<E>::type>(e);
}

then use it:

auto value = to_integral(my_fields::field);

auto redValue = to_integral(Color::Red);//where Color is an enum class!

And since the function is declared to be constexpr, you can use it where constant expression is required:

int a[to_integral(my_fields::field)]; //declaring an array

std::array<int, to_integral(my_fields::field)> b; //better!
42
votes

You cannot convert it implicitly, but an explicit cast is possible:

enum class my_fields : unsigned { field = 1 };

// ...

unsigned x = my_fields::field; // ERROR!
unsigned x = static_cast<unsigned>(my_fields::field); // OK

Also mind the fact, that the semicolon should be after the closed curly brace in your enum's definition, not before.

1
votes

As others have pointed out there is no implicit cast, but you can use an explicit static_cast. I use the following helper functions in my code to convert to and from an enum type and its underlying class.

    template<typename EnumType>
    constexpr inline decltype(auto) getIntegralEnumValue(EnumType enumValue)
    {
        static_assert(std::is_enum<EnumType>::value,"Enum type required");
        using EnumValueType = std::underlying_type_t<EnumType>;
        return static_cast<EnumValueType>(enumValue);
    }

    template<typename EnumType,typename IntegralType>
    constexpr inline EnumType toEnum(IntegralType value)
    {
        static_assert(std::is_enum<EnumType>::value,"Enum type required");
        static_assert(std::is_integral<IntegralType>::value, "Integer required");
        return static_cast<EnumType>(value);
    }

    template<typename EnumType,typename UnaryFunction>
    constexpr inline void setIntegralEnumValue(EnumType& enumValue, UnaryFunction integralWritingFunction)
    {
        // Since using reinterpret_cast on reference to underlying enum type is UB must declare underlying type value and write to it and then cast it to enum type
        // See discussion on https://stackguides.com/questions/19476818/is-it-safe-to-reinterpret-cast-an-enum-class-variable-to-a-reference-of-the-unde

        static_assert(std::is_enum<EnumType>::value,"Enum type required");

        auto enumIntegralValue = getIntegralEnumValue(enumValue);
        integralWritingFunction(enumIntegralValue);
        enumValue = toEnum<EnumType>(enumIntegralValue);
    }

Usage code

enum class MyEnum {
   first = 1,
   second
};

MyEnum myEnum = MyEnum::first;
std::cout << getIntegralEnumValue(myEnum); // prints 1

MyEnum convertedEnum = toEnum(1);

setIntegralEnumValue(convertedEnum,[](auto& integralValue) { ++integralValue; });
std::cout << getIntegralEnumValue(convertedEnum); // prints 2
0
votes

I find the following function underlying_cast useful when having to serialise enum values correctly.

namespace util
{

namespace detail
{
    template <typename E>
    using UnderlyingType = typename std::underlying_type<E>::type;

    template <typename E>
    using EnumTypesOnly = typename std::enable_if<std::is_enum<E>::value, E>::type;

}   // namespace util.detail


template <typename E, typename = detail::EnumTypesOnly<E>>
constexpr detail::UnderlyingType<E> underlying_cast(E e) {
    return static_cast<detail::UnderlyingType<E>>(e);
}

}   // namespace util

enum SomeEnum : uint16_t { A, B };

void write(SomeEnum /*e*/) {
    std::cout << "SomeEnum!\n";
}

void write(uint16_t /*v*/) {
    std::cout << "uint16_t!\n";
}

int main(int argc, char* argv[]) {
    SomeEnum e = B;
    write(util::underlying_cast(e));
    return 0;
}