dynamic_cast< D *>(pb)返回null [英] dynamic_cast<D *>(pb) return null

问题描述

在C ++入门(第5版)19.2.1中，关于dynamic_cast.它说，要使 dynamic_cast< type *>(e)成功，

In C++ primer(5th) 19.2.1 about dynamic_cast. It says, for dynamic_cast<type*>(e) to be successful,

e的类型必须是从以下对象公开派生的类类型目标类型，目标类型的公共基类或与目标相同输入

但是，对于以下代码:

class B{
  public:
    virtual ~B(){}
};

class D : public B{};

B *pb = new B;
D *pd = dynamic_cast<D*>(pb);
if(pd == 0) cout << "err" << endl;

输出为"err".但是pb的类型是D类型的公共基类.

The output is "err". But the type of pb is a public base class of type D.

这是C ++入门(第5个)中的错误吗?还是我只是误解了这些话?

Is this a mistake in C++ primer(5th)? Or do I just misunderstand these words?

推荐答案

pb 的类型确实是 D 的公共基类，但 P 的对象> pb 指向的 not 不是 D 类型的任何对象的基础子对象.动态类型转换会检测到此情况并返回null.

The type of pb is indeed a public base class of D, but the object that pb points to is not the base subobject of any object of type D. The dynamic cast detects this and returns null.

如果确实确实尝试将指针转换为 D 对象的基础子对象，则将获得(非空)指向 D 对象的指针:

If indeed you did attempt to cast the pointer to a base subobject of a D object, you would get the (non-null) pointer to the D object:

D obj;
B *pb = &obj;   // points at subobject

assert(&obj == dynamic_cast<D*>(pb));

您引用的要求仅仅是一个静态要求，它允许您完全使用动态强制转换-但是它没有描述使用强制转换的结果.稍后将对此进行描述.

The requirement that you've cited is merely a static requirement that allows you to use the dynamic cast at all -- but it does not describe the result of using the cast. That's described later on.

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