将超类函数作为非typename模板参数传递

Suppose you have

假设你有

  struct A{
  void f(){}
};


struct B:public A{
};


template<typename C,void (C::*f)()>
struct Call{

  void operator()(C* c){
    (c->*f)();
  }
};
  struct A{
  void f