使用基类指针创建对象时,缺少派生类析构函数
在下面的代码示例中,未调用派生类析构函数。知道为什么吗?
我有一个具有虚函数的基类。现在,我使用基类指针来创建派生类的新对象。我的理解是,当销毁派生类对象时,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++的理解在哪里不正确?
#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
解决方案
基类的析构函数需要是虚的:
virtual ~DomesticAnimals() {
cout << "Domestic Animal kill" << endl;
};
为了避免混淆(参见注释):只有在通过指向其基类的指针删除对象的情况下,才需要将析构函数设为虚的。当需要多态时,通常是这种情况。
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