c++ - 为每个模板实例化生成唯一的类型或 ID？ (示例观察者模式)

Question

有没有一种方法或技术可以在编译时为每个模板实例生成唯一的类型或 ID？

例如这个观察者模式:

#include <set>
#include <iostream>

template <typename T>
struct type2type {};    // maybe int2type


template<class T, class T_UNIQUE>
struct OBSERVER_BASE
{
  virtual void notify ( T, type2type< T_UNIQUE > ) = 0;
};

template<class T, class T_UNIQUE>
struct SUBJECT_BASE
{
  // This i like to do without the T_UNIQUE parameter
  typedef T_UNIQUE unique_type;

  std::set< OBSERVER_BASE< T, unique_type >* > my_observer{};

  void do_notify ()
  {
    for ( auto obs : my_observer )
      obs->notify ( T{}, type2type< unique_type >{} );
  }
};


class X {};
class Y {};
                                          // manual unique required?
class Subject_A : public SUBJECT_BASE< X, Subject_A > {};
class Subject_B : public SUBJECT_BASE< X, Subject_B > {};
class Subject_C : public SUBJECT_BASE< Y, Subject_C > {};

// typedef UNIQUE_. only to illustrate the idea 
typedef typename Subject_A::unique_type UNIQUE_A;
typedef typename Subject_B::unique_type UNIQUE_B;
typedef typename Subject_C::unique_type UNIQUE_C;

class Observer :
  public OBSERVER_BASE< X, UNIQUE_A >,
  public OBSERVER_BASE< X, UNIQUE_B >,
  public OBSERVER_BASE< Y, UNIQUE_C >
{
  virtual void notify ( X, type2type< UNIQUE_A > ) override
  {
    std::cout << "x from Subject_A" << std::endl;
  }

  virtual void notify ( X, type2type< UNIQUE_B > ) override
  {
    std::cout << "x from Subject_B" << std::endl;
  }

  virtual void notify ( Y, type2type< UNIQUE_C > ) override
  {
    std::cout << "y from Subject_C" << std::endl;
  }
};

int main ( int argc, char **argv )
{
  Subject_A sub_a {};
  Subject_B sub_b {};
  Subject_C sub_c {};

  Observer obs {};

  sub_a.my_observer.insert( &obs );
  sub_b.my_observer.insert( &obs );
  sub_c.my_observer.insert( &obs );

  sub_a.do_notify();
  sub_b.do_notify();
  sub_c.do_notify();
}

有没有办法以这种方式做到这一点(无需手动唯一参数)？我知道听起来很奇怪...

template<class T>
struct SUBJECT_BASE
{
  typedef AUTOMATIC_UNIQUE_TYPE__OR__WHAT_EVER unique_type;
};

class Subject_A : public SUBJECT_BASE< X > {};
class Subject_B : public SUBJECT_BASE< X > {};
class Subject_C : public SUBJECT_BASE< Y > {};

Answer 1

我认为没有办法实现您的要求，但有可能实现您似乎想要的。此实现不需要手动 UNIQUE_TYPE:

template<class T>
class Subject
{
  std::vector<std::function<void(T)>> observers;

public:
  void do_notify () {
    for ( auto& obs : observers )
      obs( T{} );
  }

  void add_listener(std::function<void(T)> l) {
      observers.emplace_back(std::move(l));
  }
};

class X {};
class Y {};

int main ()
{
  Subject<X> sub_a {};
  Subject<X> sub_b {};
  Subject<Y> sub_c {};

  sub_a.add_listener([](X){std::cout << "x from Subject_A" << std::endl;});
  sub_b.add_listener([](X){std::cout << "x from Subject_B" << std::endl;});
  sub_c.add_listener([](Y){std::cout << "y from Subject_C" << std::endl;});

  sub_a.do_notify();
  sub_b.do_notify();
  sub_c.do_notify();
}

但它的 OOP-ey 比您发布的代码要少得多。 ( Live at Coliru )