在没有虚析构函数的情况下删除多态对象会发生什么？
What happens when delete a polymorphic object without a virtual destructor?
在下面的示例中，b 是一个多态指针类型，其 static 类型为 Base* 且其 dynamic 类型是 Derived*.
struct Base
{
virtual void f();
};
struct Derived : Base
{
};
int main()
{
Base *b = new Derived();
// ...
delete b;
}
在没有虚拟析构函数的情况下删除 b 会发生什么？
What happens when b is deleted without a virtual destructor?
我们不知道。行为未定义。对于大多数实际情况，Derived 的析构函数可能不会被调用，但没有任何保证。
（强调我的）
In the first alternative (delete object), if the static type of the
object to be deleted is different from its dynamic type, the static
type shall be a base class of the dynamic type of the object to be
deleted and the static type shall have a virtual destructor or the
behavior is undefined.
在下面的示例中，b 是一个多态指针类型，其 static 类型为 Base* 且其 dynamic 类型是 Derived*.
struct Base
{
virtual void f();
};
struct Derived : Base
{
};
int main()
{
Base *b = new Derived();
// ...
delete b;
}
在没有虚拟析构函数的情况下删除 b 会发生什么？
What happens when b is deleted without a virtual destructor?
我们不知道。行为未定义。对于大多数实际情况，Derived 的析构函数可能不会被调用，但没有任何保证。
（强调我的）
In the first alternative (delete object), if the static type of the object to be deleted is different from its dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the static type shall have a virtual destructor or the behavior is undefined.