Question

据我所知，如果我们想调用派生对象的析构函数，该派生对象已分配给指向base的指针，我们希望将基本析构函数设为虚拟。但是，如果我们有这样的事情：

#include <iostream>
using namespace std;
class base
{

public:
base() { cout << "Base Constructor Called\n"; }
virtual ~base() { cout << "Base Destructor called\n"; }

};
class derived1 :public base
{

public:
derived1() { cout << "Derived1 constructor called\n"; }
~derived1() { cout << "Derived1 destructor called\n"; }

 };

 class derived2 : public derived1
 {
public:
derived2() { cout << "Derived2 constructor called\n"; }
~derived2() { cout << "Derived2 destructor called\n"; }

  };

class derived3 : public derived2
{
  public:
derived3() { cout << "Derived3 constructor called\n"; }
~derived3() { cout << "Derived3 destructor called\n"; }

 };

我们有这样的主要功能：

int main (){
base* ptr=new derived3;
delete ptr;

输出为：
Base Constructor Called Derived1 constructor called Derived2 constructor called Derived3 constructor called Derived3 destructor called Derived2 destructor called Derived1 destructor called Base Destructor called

这会调用base，derived1，derived2和derived3析构函数，它们运行得很好。我们只将基础析构函数设为虚拟为什么没有必要将derived1和derived2析构函数作为虚拟来产生相同的结果？

Answer 1

从具有虚析构函数的类派生的所有类都隐式具有虚拟析构函数。

这适用于类层次结构中的任何成员函数。

Answer 2

将函数声明为virtual后，它在所有派生类中隐式保留virtual。

虚拟析构函数C ++

2 个答案: