提供特征 class 的类型的模板特化

Question

I have a class

template <typename T> struct Dispatch;

which is used to call a type-specific function. For instance, assume I have dispatchers like

template <> struct Dispatch <MyClass> {
  static void Apply (void* a, MyClass& m)
  {
      ::memcpy (a, &m, sizeof (m));
  }
};

Now I have a bunch of classes for which I have a type-trait, ArrayTypes . I would like to do something like:

 template <> struct Dispatch <enable_if<IsArrayType>>
 {
   template <typename ArrayType>
   static void Apply (void* a, ArrayType& m)
   {
     ::memcpy (a, &m, ArrayTypeTraits<ArrayType>::GetSize (m));
   }
 };

Is this possible?

Answer 1

Use boost enable_if .

If boost is unavailable, check out the enable_if idiom .

Answer 2

Just found it:

template <typename T, class Enable = void> struct Dispatch;
template <typename T>
struct Dispatch<T, typename boost::enable_if< typename IsArrayType<T>::type>::type>
{
};

Thanks to Kornel.

Answer 3

As of C++14 this is standardized as std::enable_if .

Answer 4

The idea behind traits is that you can provide a default implementation, and types can optionally specialize that. This way, you have no special cases in the code that uses the traits. Your approach (having special cases for classes with and without traits defined) defeats the purpose of type-traits.