c++ - 动态转换规范(规则)说明

Question

我们有 dynamic_cast 的一般形式:

dynamic_cast < new-type > ( expression )

我对这条规则 (5a) 的粗体部分特别困惑:

5: If expression is a pointer or reference to a polymorphic type Base,and new-type is a pointer or reference to the type Derived a run-timecheck is performed:

a) The most derived object pointed/identified by expression isexamined. If, in that object, expression points/refers to a publicbase of Derived, and if only one object of Derived type is derivedfrom the subobject pointed/identified by expression, then the resultof the cast points/refers to that Derived object. (This is known as a"downcast".)

你能举一个例子，这部分不满足吗？
以上摘录来自 cppreference: cppreferenc

Answer 1

充实多继承示例@Peter 总结:

     Base1
     /   \  <-- Virtual inheritance here
  Base2  Base2
    |     | <-- Nonvirtual inheritance here and below
  Left   Right
    \     /
    Derived

Base1* p_base1 = new Derived();
Base2* p_base2 = dynamic_cast<Base2*>(p_base1); // Which Base2?

有两种不同的 Base2 Derived 中的对象对象，那么应该是哪个 p_base2指向？
代码示例:

#include <iostream>

struct Base1 { virtual ~Base1() = default; };
struct Base2 : virtual Base1 { };
struct Left : Base2 { };
struct Right : Base2 { };
struct Derived : Left, Right {
    Derived() : Base1() {}
};

int main()
{
    Base1* p_base1 = new Derived();
    Base2* p_base2 = dynamic_cast<Base2*>(p_base1);
    std::cout << std::boolalpha;
    std::cout << "p_base1 == nullptr: " << (p_base1 == nullptr) << '\n';
    std::cout << "p_base2 == nullptr: " << (p_base2 == nullptr);
    delete p_base1;
}

在这里要小心一点: Base1几乎是继承的，所以只有一个 Base1子对象，我们实际上可以初始化 p_base1 .然而， Derived非虚拟地继承自 Left和 Right ，这意味着它有两个 Base2 的实例.因此，向下转型失败。
输出:

p_base1 == nullptr: false
p_base2 == nullptr: true