Template specialization for enum

Question

Is it possible to specialize a templatized method for enums?

Something like (the invalid code below):

template <typename T>
void f(T value);

template <>
void f<enum T>(T value);

In the case it's not possible, then supposing I have specializations for a number of types, like int, unsigned int, long long, unsigned long long, etc, then which of the specializations an enum value will use?

Answer 1

You can use std::enable_if with std::is_enum from <type_traits> to accomplish this.

In an answer to one of my questions, litb posted a very detailed and well-written explanation of how this can be done with the Boost equivalents.

Answer 2

I'm not sure if I understand your question correctly, but you can instantiate the template on specific enums:

template <typename T>
void f(T value);

enum cars { ford, volvo, saab, subaru, toyota };
enum colors { red, black, green, blue };

template <>
void f<cars>(cars) { }

template <>
void f<colors>(colors) { }

int main() {
    f(ford);
    f(red);
}

Answer 3

Presumably, the only interesting thing you could do with a type that they only thing you know about it is that it's an enum, is cast it to its underlying type and operate on that. Here's how that might look like, using James' suggested approach (AKA SFINAE):

void Bar(int b); // and/or other underlying types

template<typename T>
typename std::enable_if<std::is_enum<T>::value, void>::type
Foo(T enm)
{
    Bar(static_cast<typename std::underlying_type<T>::type>(enm));
}

As a related bonus, here's a similar method that would only get resolved for a specific type of your choosing (replace bool in is_same to the type of your choosing):

template<typename T>
typename std::enable_if<std::is_same<T,bool>::value, void>::type
Baz(T bl)
{
    if (bl)
    {
        //...
    }
    else
    {
        //...
    }
}