不能使用列舉類作為unordered_map鍵

I use a functor object to calculate hash of enum class:

struct EnumClassHash
{
    template <typename T>
    std::size_t operator()(T t) const
    {
        return static_cast<std::size_t>(t);
    }
};

Now you can use it as 3rd template-parameter of std::unordered_map:

enum class MyEnum {};

std::unordered_map<MyEnum, int, EnumClassHash> myMap;

So you don't need to provide a specialization of std::hash, the template argument deduction do the job. Furthermore, you can use the word using and make your own unordered_map that use std::hash or EnumClassHash depending on the Key type:

template <typename Key>
using HashType = typename std::conditional<std::is_enum<Key>::value, EnumClassHash, std::hash<Key>>::type;

template <typename Key, typename T>
using MyUnorderedMap = std::unordered_map<Key, T, HashType<Key>>;

Now you can use MyUnorderedMap with enum class or another type:

MyUnorderedMap<int, int> myMap2;
MyUnorderedMap<MyEnum, int> myMap3;

Theoretically, HashType could use std::underlying_type and then the EnumClassHash will not be necessary. That could be something like this, but I haven't tried yet:

template <typename Key>
using HashType = typename std::conditional<std::is_enum<Key>::value, std::hash<std::underlying_type<Key>::type>, std::hash<Key>>::type;

If using std::underlying_type works, could be a very good proposal for the standard.