C++学习笔记25,析构函数总是会宣布virtual

为了永远记住析构函数声明virtual----><<effective c++>>

为这句话不一定对,但无需质疑的是这句话是非常实用的.

查看以下的样例:

#include <iostream>
#include <string>
using namespace std;
class B{
public:
	~B(){
		cout<<"base is destroyed!"<<endl;
	}
};
class D:public B{
public:
	~D(){
		cout<<"dirved is destroyed!"<<endl;
	}

};

int main(){
	{
	D d;
	}
}

这个是一个非常easy的样例,输出是肯定的.

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可是,再看以下的測试!

#include <iostream>
#include <string>
using namespace std;
class B{
public:
	~B(){
		cout<<"base is destroyed!"<<endl;
	}
};
class D:public B{
public:
	~D(){
		cout<<"dirved is destroyed!"<<endl;
	}

};

int main(){
	{
	B *b=new D();
	delete b;
	}
}

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因此能够看到,当用一个基类的指针指向一个派生类的对象,直接删除该指针时,析构函数调用链将会遭到破坏.

#include <iostream>
#include <string>
using namespace std;
class B{
public:
	virtual ~B(){
		cout<<"base is destroyed!"<<endl;
	}
};
class D:public B{
public:
	virtual ~D(){
		cout<<"dirved is destroyed!"<<endl;
	}

};

int main(){
	{
	B *b=new D();
	delete b;
	}
}

因此应该记得将析构函数声明为virtual,这样就永远不用操心析构函数调用链的破坏!

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