为两种不同类型专门化相同的模板定义

Question

I am trying to specialize the same template for two different classes in the following way:

template <>
struct TemplateClass<A> {
  void method(A x) {
    std::cout << x << std::endl;
  }
};

template <>
struct TemplateClass<B> {
  void method(B x) {
    std::cout << x << std::endl;
  }
};

Is there a way I can rewrite this to avoid the duplication inside the two specialization definitions?

I do not have control over the definition of TemplateClass . The particular application I have in mind is specialization of fmt::formatter as shown here: https://fmt.dev/latest/api.html#formatting-user-defined-types

Answer 1

If you look at the definition of struct formatter in fmt/core.h , it has a dummy typename = void parameter:

template <typename T, typename Char = char, typename Enable = void>
struct formatter

This dummy parameter allows you to do SFINAE in specializations:

template <typename T>
struct formatter<T, char, std::enable_if_t<std::is_same_v<T, A> || std::is_same_v<T, B>>>
{
    // ...
};

Answer 2

The only duplication you could avoid in your presented code is the std::cout << std::endl; You are outputting 'x' in both but they are of two different types so that is hardly duplication. If the duplication were actually enough to be concerned over you could create a shared base and have both specializations derive from that.