虚拟析构函数如何工作？ [英] How do virtual destructors work?

问题描述

我使用gcc。我知道如何虚拟析构函数解决问题时，我们销毁一个派生类对象指向一个基类指针。我想知道他们如何工作？

I am using gcc. I am aware how the virtual destructors solve the problem when we destroy a derived class object pointed by a base class pointer. I want to know how do they work?

class A
{
      public:
      A(){cout<<"A constructor"<<endl;}
     ~A(){cout<<"A destructor"<<endl;}

};

class B:public A
{
      public:
      B(){cout<<"B constructor"<<endl;}
      ~B(){cout<<"B destructor"<<endl;}
};       

int main()
{
  A * a = new B();
  delete a;    
  getch();
  return 0;   
} 

当我将A的析构函数更改为虚函数时，问题就解决了。这是什么内在的工作。为什么我使A的析构函数虚拟。我想知道A和B的vtable发生了什么？

When I change A's destructor to a virtual function, the problem is solved. What is the inner working for this. Why do I make A's destructor virtual. I want to know what happens to the vtable of A and B?

推荐答案

你需要知道的关键是，在上面的代码中的虚拟析构函数是未定义的行为，这不是你想要的。虚拟析构函数与任何其他虚拟函数一样 - 当你调用 delete 时，程序将决定在运行时调用哪个析构函数并解决你的问题。

The key thing you need to know is that not using a virtual destructor in the above code is undefined behavior and that's not what you want. Virtual destructors are like any other virtual functions - when you call delete the program will decide what destructor to call right in runtime and that solves your problem.

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