模板：类专业化

Question

I'm new in the C++ world. Sorry for my nooby question.

I have a class

template <typename T>
class Foo
{
  T t_;
  void say_hello() 
  { std::cout << "Ciao";}
  // work with T ...
};

I want to specialize this template class for 3 types.

If type is ( A or B or C ), Then use this class

template<>
class Foo<A or B or C>
{
  void say_hello() 
  { std::cout << "Hello";}
};

What's the best way to do this? Thank you for your help.

Answer 1

A possible solution uses SFINAE

template <typename T, typename = void>
class Foo
{
  T t_;
  void say_hello() 
  { std::cout << "Ciao";}
  // work with T ...
};

template <typename T>
class Foo<T, std::enable_if_t<std::is_same_v<T, A>,
                           || std::is_same_v<T, B>,
                           || std::is_same_v<T, C>>
{
  void say_hello() 
  { std::cout << "Hello";}
};

If you don't use T inside the Foo specialization (as in your example) you can also use a sort of self-inheritance

template <typename T>
class Foo
{
  T t_;
  void say_hello() 
  { std::cout << "Ciao";}
  // work with T ...
};

template <>
class Foo<A>
{
  void say_hello() 
  { std::cout << "Hello";}
};

template <>
class Foo<B> : public Foo<A>
 { };

template <>
class Foo<C> : public Foo<A>
 { };

Off Topic: if you want to use say_hello() outside the class, is better if you make it public (or if you declare Foo as a struct ).

Answer 2

There are several possibilities, for example:

Specialization of the method only:

template<>
void Foo<A>::say_hello() { std::cout << "Hello"; }
template<>
void Foo<B>::say_hello() { std::cout << "Hello"; }
template<>
void Foo<C>::say_hello() { std::cout << "Hello"; }

or, in C++17, you might do:

template <typename T>
class Foo
{
  T t_;
  void say_hello() 
  {
      if constexpr(std::is_same_v<T, A> || std::is_same_v<T, B> || std::is_same_v<T, C>) {
          std::cout << "Hello";
      } else {
          std::cout << "Ciao";
      }
  }
  // work with T ...
};

Whereas regular if works in that example, it would fail if you call code specific to A , B , C . if constexpr won't have that issue.

Answer 3

A variant of the SFINAE solution that is a bit more concise for more classes.

template<class T, class... Ts>
struct is_one_of;

template<class T, class Ts>
struct is_one_of<T, T, Ts...> : std::true_type {}; //maybe add std::decay_t

template<class T, class S, class Ts>
struct is_one_of<T, S, Ts...> : is_one_of<T, Ts...> {};

template<class T>
struct is_one_of<T> : std::false_type{};

template<class T, class... Ts>
constexpr bool is_one_of_v = is_one_of<T, Ts...>::value;

template <typename T, typename = void>
class Foo
{
  T t_;
  void say_hello() 
  { std::cout << "Ciao";}
  // work with T ...
};

template <typename T>
class Foo<T, std::enable_if_t<is_one_of_v<T, A, B, C>
{
  void say_hello() 
  { std::cout << "Hello";}
};