Can a template parameter's default argument be specialized?

Question

In C++, if I have a template parameter, how can I cleanly specialize a default argument? For example, consider the following:

template <class Key, class Value = int > class Association;

What if I want Value to instead default to float for class Special? Is there a way to in effect specialize the class Association such that if Key is Special that Value defaults to instead be float?

I imagine one way to do this would be with traits:

template <class Key> struct Traits {
  typedef int defaultValue;
}
template<> struct Traits<Special> {
  typedef float defaultValue;
}
template <class Key, class Value = Traits<Key>::defaultValue> class Association;

Is there a more succinct way of doing this that is not so involved and would more readily show that int is the normal default at the place where Association is defined?

Answer 1

Well, a not-necessarily-prettier one-liner:

#include <type_traits>

template <typename Key,
          typename Value = typename std::conditional<std::is_same<Key, Special>::value, float, int>::type>
class Association { /* ... */ };