相关疑难解决方法(0)

以下代码如何工作？

typedef char (&yes)[1];
typedef char (&no)[2];

template <typename B, typename D>
struct Host
{
  operator B*() const;
  operator D*();
};

template <typename B, typename D>
struct is_base_of
{
  template <typename T> 
  static yes check(D*, T);
  static no check(B*, int);

  static const bool value = sizeof(check(Host<B,D>(), int())) == sizeof(yes);
};

//Test sample
class Base {};
class Derived : private Base {};

//Expression is true.
int test[is_base_of<Base,Derived>::value && !is_base_of<Derived,Base>::value];

1. 请注意,这B是私人基地.这是如何运作的？

2. 注意operator B*()是const.它为什么如此重要？

3. 为什么template<typename T> static yes …

在/sf/answers/137702841/中,提供了一种解决方案,用于静态检查成员是否存在,可能在类型的子类中:

template <typename Type> 
class has_resize_method
{ 
   class yes { char m;}; 
   class no { yes m[2];}; 
   struct BaseMixin 
   { 
     void resize(int){} 
   }; 
   struct Base : public Type, public BaseMixin {}; 
   template <typename T, T t>  class Helper{}; 
   template <typename U> 
   static no deduce(U*, Helper<void (BaseMixin::*)(), &U::foo>* = 0); 
   static yes deduce(...); 
public: 
   static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0))); 
};

但是,它不适用于C++ 11 final类,因为它继承了被测试的类,这会final阻止它.

OTOH,这一个:

template <typename C>
struct has_reserve_method {
private:
    struct No …