使用基类指针创建对象时，缺少派生类析构函数

Question

在下面的代码示例中，没有调用派生类析构函数。知道为什么吗？我有一个基类，它具有虚拟函数。现在我使用基类指针来创建派生类的新对象。我的理解是，当派生类对象被销毁时，首先调用派生类的析构函数，然后调用基类。然而，我只看到基类的析构函数被调用。有人知道我做错了什么吗？或者我对c++的理解哪里不正确？

#include <iostream>
#include <bitset>
using namespace std; 
class DomesticAnimals 
{
public:
DomesticAnimals() {
cout <<"Domestic Animal Base" <<endl;
}; 
~DomesticAnimals() {
cout <<"Domestic Animal kill " <<endl;
}; 
virtual void Speak() = 0;
virtual void EatFood()= 0; 
virtual int NoOfLegs() {
return 4; 
} ;
};

class Cat : public DomesticAnimals
{
public:
Cat(); 
~Cat(); 
void Speak() override; 
void EatFood() override; 
};
Cat::Cat()
{
cout << "Kat was born" << endl;
}
Cat:: ~Cat()
{
cout << "Kill de Cat" << endl; 
}
void Cat:: Speak()
{
cout << "Meow Meow " << endl; 
}
void Cat::EatFood()
{
cout <<"Kat eet de muis vandaag !! " <<endl; 
}

class Dog : public DomesticAnimals
{
public:
Dog();
~Dog();
void Speak() override; 
void EatFood() override; 
};
Dog::Dog()
{
cout << "A puppy was born" << endl; 
}
Dog::~Dog()
{
cout << "Kill de hond" << endl; 
}
void Dog :: Speak()
{
cout <<"bow bow woof woof" <<endl;
}
void Dog :: EatFood()
{
cout << "de hond eet een kip voor middageten" <<endl;
}
void DogActions()
{
DomesticAnimals* dog = new Dog;
cout<< endl;
dog->Speak(); 
dog->EatFood();
cout<<"No of legs for dog = "<< dog->NoOfLegs() <<endl; 
cout<<endl;
delete dog; 
dog = NULL;
}
void CatActions()
{
DomesticAnimals* cat = new Cat; 
cat->Speak(); 
cat->EatFood(); 
cout<<"No of legs for cat = "<< cat->NoOfLegs() << endl;
delete cat; 
cat = NULL; 
}
int main(void)
{
DogActions();
CatActions();
return 0;
}

程序的输出如下

Domestic Animal Base
A puppy was born
bow bow woof woof
de hond eet een kip voor middageten
No of legs for dog = 4
Domestic Animal kill 
Domestic Animal Base
Kat was born
Meow Meow 
Kat eet de muis vandaag !! 
No of legs for cat = 4
Domestic Animal kill 

Answer 1

基类的析构函数需要是虚拟的：

virtual ~DomesticAnimals() {
cout << "Domestic Animal kill" << endl;
};

为了避免任何混淆(请参阅注释(：只有在通过指向基类的指针删除对象的情况下，才需要使析构函数成为虚拟的。当需要多态性时，通常会出现这种情况。