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.
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;
}
}
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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