今天在一场面试过程中碰到这个问题,当时一时片刻没有反应过来,一头雾水。只记得自己回到说,是会造成内存泄露,但面试穷追猛打,一直追问为什么造成内存泄露,还给举例说明,一般情况下是不会造成内存泄露的,搞得场面很尴尬。回来后,心有不甘,上机看看到底存在不存在这个问题。
#include <iostream>using namespace std;class Base{public:Base(){cout<<"Base::Base"<<endl;};~Base(){cout<<"Base::~Base"<<endl;} };class Derived : public Base{public:Derived():Base(){cout<<"Derived::Derived"<<endl;buf = new char[1024];}~Derived() {cout<<"Derived::~Derived"<<endl;delete[] buf; }private:char* buf;};int main(){Base* a = new Derived();delete a;return 0;}
运行结果
Base::BaseDerived::DerivedBase::~Base
果然没有释放掉继承类的析构函数,好了。当我把基类的析构函数改为virtual的情况又如何呢?
#include <iostream>using namespace std;class Base{public:Base(){cout<<"Base::Base"<<endl;};virtual ~Base(){cout<<"Base::~Base"<<endl;} };class Derived : public Base{public:Derived():Base(){cout<<"Derived::Derived"<<endl;buf = new char[1024];}~Derived() {cout<<"Derived::~Derived"<<endl;delete[] buf; }private:char* buf;};int main(){Base* a = new Derived();delete a;return 0;}
运行结果
Base::BaseDerived::DerivedDerived::~DerivedBase::~Base
结论
为什么会是这样呢? 先补充一个知识, 派生类构造函数和析构函数都会调用基类的构造函数和析构函数,而且顺序是基类的构造函数先调用,析构的时候基类的析构函数最后调用,从上面例子输出中也能看出来。 还有最重要的一点就是virtual就是为多态准备的,只有加上virtual的声明的函数,才能实现多态特性,对普通函数我们很好理解,但对于特殊的函数比如析构函数,也是一样的,只不过它不是同名函数,但具有的性质是一样的。