c++ - dynamic_casting 对象混淆

Question

本书The c++ programming language有关于 dynamic_cast 的部分，我不确定我是否理解正确。

The purpose of dynamic_cast is to deal with the case in which the correctness of the conversion cannot be determined by the compiler. In that case, dynamic_cast(p) looks at the object pointed to by p (if any). If that object is of class T or has a unique base class of type T, then dynamic_cast returns a pointer of type T* to that object; otherwise, nullptr is returned. If the value of p is nullptr, dynamic_cast(p) returns nullptr. Note the requirement that the conversion must be to a uniquely identified object. It is possible to construct examples where the conversion fails and nullptr is returned because the object pointed to by p has more than one subobject representing bases of type T.

“是否可以构建转换失败并返回 nullptr 的示例，因为对象指向 to by p has more than one subobject representing bases of type T"意思是这样的？

class a {
public:
    a() { }
};

class b : public a {
public:
    b() { }
};

class z : public a, public b {
public:
    z() { }
};

void f(z* p) {
    a* x = dynamic_cast<a*>(p); // ambiguous
}

还有一个，这是摘自书本:

class Component : public virtual Storable { /* ... */ };
class Receiver : public Component { /* ... */ };
class Transmitter : public Component { /* ... */ };
class Radio : public Receiver, public Transmitter { /* ... */ };

The ambiguity for a pointer to a Radio object is not in general detectable at compile time. This kind of run-time ambiguity detection is needed only for virtual bases. For ordinary bases, there is always a unique subobject of a given cast (or none) when downcasting (that is, toward a derived class; §22.2). The equivalent ambiguity for virtual bases occurs when upcasting (that is, toward a base), but such ambiguities are caught at compile time.

我完全不明白这个。这是什么意思“对于普通基地，给定 Actor 总是有一个独特的子对象”？我知道如果基不是虚拟的，将为从它派生的每个类创建一个子对象。但就转换而言，我只是在上面的例子中造成了歧义错误。而“向上转换时会发生虚拟基础的等效歧义”，这是什么意思？虚拟基地可以模棱两可吗？谁能解释得更清楚？

Answer 1

” For ordinary [non-virtual] bases, there is always a unique subobject of a given cast (or none) when downcasting (that is, toward a derived class; §22.2

那是因为非虚拟继承创建了派生类的严格层次结构，分支不会再次聚集在一起。对象必须属于某个最派生的类 T，位于该类树的叶节点下方。往上走，就不能再遇到T了，因为一个类是不能继承自己的。因此，通过动态向下转换为 T，您最终将得到对象的最派生类。对于(或每个)继承链中的任何类也是如此。