如何访问受保护的基类函数，从派生类通过基类ptr

问题描述

我有抽象类A，我从中继承了一些类。在派生类中，我试图访问A槽A指针中的保护函数。但我得到一个编译器错误。

I have abstract class A, from which I inherit a number of classes. In the derived classes I am trying to access protected function in A trough A pointer. But I get a compiler error.

class A
{
   protected:
        virtual  void f()=0;
};

class D : public A
{
    public:
         D(A* aa) :mAPtr(aa){}
         void g();

    protected:
         virtual void f();

    private:
      A* mAPtr; // ptr shows to some derived class instance
};

void D::f(){  }


void D::g()
{
   mAPtr->f();
}

编译器错误说：无法访问在类中声明的受保护成员A :: f A。

The compiler error says : cannot access protected member A::f declared in class A.

如果我声明mAPtr为D *，而不是A *所有的编译。

If I declare mAPtr to be D*, instead A* everything compiles. And I don't understand why is this.

推荐答案

依赖私人访问对同一类型的不相关实例的访问。

Relying on private access works on unrelated instances of the same type.

依赖 protected

但是，依赖 protected 访问 em>工作于基本类型的无关实例。

However, relying on protected access does not work on unrelated instances of a base type.

所以 D 或派生自 D 的东西， code> A 。

So D or something derived from D, but not A.

这是一个可疑的奇怪的C ++，但是设计，以避免陷阱。毕竟，你不知道 * mAPtr 真的类型。

It's an oft-questioned cute oddity about C++ that nonetheless is designed to try to avoid pitfalls. After all, you don't know what type *mAPtr really has.

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