转:析构函数
析构函数在下边3种情况时被调用: 1.对象生命周期结束,被销毁时; 2.delete指向对象的指针时,或delete指向对象的基类类型指针,而其基类虚构函数是虚函数时; 3.对象i是对象o的成员,o的析构函数被调用时,对象i的析构函数也被调用。
情况1请看下边代码:
#include<iostream.h>
class A {
public: A() { cout<<"constructing A"<<endl; }
~A() { cout<<"destructing A"<<endl; }
private: int a;
};
class B: public A {
public: B() { cout<<"constructing B"<<endl; }
~B() { cout<<"destructing B"<<endl; }
private: int b;
}; void main() {
B b;
}
运行结果为:
constructing A
constructing B
destructing B
destructing A
上述代码还说明了一件事:析构函数的调用顺序与构造函数的调用顺序相反。
情况2则正好说明了为什么基类应该把析构函数声明为虚函数,请先看下边的例子:
#include<iostream.h> class A { public: A() { cout<<"constructing A"<<endl; } ~A() { cout<<"destructing A"<<endl; } private: int a; }; class B: public A { public: B() { cout<<"constructing B"<<endl; } ~B() { cout<<"destructing B"<<endl; } private: int b; }; void main() { A* a = new B; delete a; }
运行结果为:
constructing A
constructing B
destructing A
若将class A中的析构函数声明为虚函数,运行结果将变成:
constructing A constructing B destructing B destructing A
由此还可以看出虚函数还是多态的基础,才c++中没有虚函数就无法实现多态。因为不声明成虚函数就不能“推迟联编”,所以不能实现多态。这点上和java不同,java总是“推迟联编”的,所以也剩了这些麻烦。
扯远了,再看情况3,通过下边代码表示:
#include<iostream.h> class A { public: A() { cout<<"constructing A"<<endl; } ~A() { cout<<"destructing A"<<endl; } private: int a; }; class C { public: C() { cout<<"constructing C"<<endl; } ~C() { cout<<"destructing C"<<endl; } private: int c; }; class B: public A { public: B() { cout<<"constructing B"<<endl; } ~B() { cout<<"destructing B"<<endl; } private: int b; C c; }; void main() { B b; }
运行结果为:
constructing A
constructing C
constructing B
destructing B
destructing C
destructing A
b的析构函数调用之后,又调用了b的成员c的析构函数,同时再次验证了析构函数的调用顺序与构造函数的调用顺序相反。
若将上边的代码中的main()函数内容改成
A* a = new B; delete a;
由情况2我们知道,这将不会调用class B的析构函数不会被调用,所以class C的析构函数也不会被调用。 正如我们想的,运行结果为:
constructing A
constructing C
constructing B
destructing A