为什么保护继承导致dynamic_cast失败？ [英] Why does protected inheritance cause dynamic_cast to fail?

问题描述

我改变了我的C ++基类为 protected 继承和我的 dynamic_cast p>

为什么要将继承更改为 protected 更改 dynamic_cast ？

  struct Base {
 static Base * lookupDerived //实际返回一个Derived *对象。 
}; 
 
 struct Derived：protected / *切换到public以使其工作* / Base {
 static void test（）{
 Base * base = lookupDerived（）; 
 
 if（dynamic_cast< Derived *>（base））{
 std :: cout< 它工作（我们必须使用公共继承）。 << std :: endl; 
} else {
 std :: cout<< 它失败了（我们必须使用受保护的继承）。 << std :: endl; 
} 
}; 
  

解决方案

您作为外部用户无法访问受保护或私人类的成员。这同样适用于受保护或私有继承。一个类的作者不希望外部用户访问受保护/私有父类超过他们希望外部用户访问受保护/私有成员。

一个原因：假设父类有一个非虚拟析构函数。从基类指针中删除实例派生类将导致未定义的行为。使父类受保护/私有意味着你不能这样做（见脚注）。

另一个原因：假设有问题的类的作者不想要外部用户有权访问父类的公共成员。可以使用公共继承（ is-a ）并将这些公共接口降级为受保护或私有，但这将违反Liskov替换原则。受保护或私有继承不是一种关系。这些公共方法变得受保护或私有，受保护或私有继承。 Liskov替换没有问题，因为protected / private继承不是

。

脚注：这里有一个丑陋的方法：Use C风格铸造。外部用户可以将
派生类指针转换为基类指针，即使基类不可访问。对我来说，这是使用 -Wold-style-cast -Werror 编译的另一个原因。

I changed my C++ base class to be protected inheritance and my dynamic_cast(s) stopped working.

Why should changing the inheritance to protected change the behavior of dynamic_cast?

struct Base {
  static Base *lookupDerived(); // Actually returns a Derived * object.
};

struct Derived : protected /* Switch this to public to get it working */ Base {
 static void test() {
   Base *base = lookupDerived();

   if (dynamic_cast<Derived *>(base)) {
      std::cout << "It worked (we must be using public inheritance)." << std::endl;
   } else {
      std::cout << "It failed (we must be using protected inheritance)." << std::endl;
   }
};

解决方案

You as an outside user cannot access the protected or private members of a class. The same applies to protected or private inheritance. The authors of a class don't want outside users to access protected/private parent classes any more than they want outside users to access protected/private members.

One reason: Suppose the parent class has a non-virtual destructor. Deleting from an instance derived class from a base class pointer would result in undefined behavior. Making the parent class protected/private means you can't do this (see footnote).

Another reason: Suppose the authors of the class in question don't want outside users to have access to the public members of the parent class. One could use public inheritance (is-a) and demote those public interfaces to protected or private, but this would violate the Liskov substitution principle. Protected or private inheritance is not an is-a relationship. Those public methods become protected or private with protected or private inheritance. There's no problem with Liskov substitution because protected/private inheritance is not is-a.

Footnote: There is an ugly way around this: Use C-style casts. Outside users can cast a derived class pointer to a base class pointer, even if the base class is not accessible. To me, that's yet another reason to compile with -Wold-style-cast -Werror.

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