CODEWITHSUNDEEP
c++enumsoperator-overloading
Kangjun Heo
Kangjun Heo

Reputation: 1073

templated operator for strongly typed enumerations

I have two enumerations with underlying type uint32_t. and I need constexpr OR operator for them because some of members should consist with result of OR of other members. I wanted to avoid to implement the operator twice or more times.

So, I tried to use template to achieve it.

    template <enum class E>
    constexpr uint32_t operator|(const E&   left, const E& right)
    {
        return (uint32_t)left | (uint32_t)right;
    }

    template <enum class E>
    constexpr uint32_t operator|(const uint32_t& left, const E& right)
    {
        return left | (uint32_t)right;
    }

    enum class U32Enum1 : uint32_t { 
        OtherMember1 = 0x01L,
        OtherMember2 = 0x02L,
        SomeOfMember = OtherMember1 | OtherMember2 // Not work
    }

    enum class U32Enum2 : uint32_t {
        OtherMember1 = 0x01L,
        OtherMember2 = 0x02L,
        SomeOfMember = OtherMember1 | OtherMember2 // Not work    
    }

However, It won't work and compiler gave me a message that no matching operands for operator |. How can I accomplish this?

Upvotes: 0

Views: 55

Answers (1)

Jarod42
Jarod42

Reputation: 217225

template <enum class E> is not the good syntax, you might do instead:

template <class E, std::enable_if_t<std::is_enum<E>::value, int> = 0>
constexpr E operator|(const E&   left, const E& right)
{
    return static_cast<E>((uint32_t)left | (uint32_t)right);
}

and using underlying type is even better:

template <class E, std::enable_if_t<std::is_enum<E>::value, int> = 0>
constexpr E operator|(const E&   left, const E& right)
{
    using T = typename std::underlying_type<E>::type;
    return static_cast<E>(static_cast<T>(left) | static_cast<T>(right));
}

Upvotes: 3

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