Virtual destruct(虚析构函数)

Virtual destruct(虚析构函数)

面试的时候,我答了虚函数是怎么工作的,但是当面试官问我为什么析构函数的基类一定需要virtual时候,我答的不是太好,所以来总结下
先说结论,当基类的指针指向子类的对象时,如果delete此指针,如果基类没有virtual析构函数,那么调用时只会调用基类的析构函数,而子类的空间未释放。
如以下例子


class BaseClass
{
private:
	int *m_Basepointer;
public:
	BaseClass() { std::cout << "it's Base Construct!" << std::endl; }
	~BaseClass()
	{
		delete m_Basepointer;
		std::cout << "it's Base Deconstruct!" << std::endl;
	}
};

class SubClass:public BaseClass
{
public:
	SubClass() { std::cout << "it's Sub Construct!" << std::endl; }
	~SubClass() 
	{
		std::cout << "it's Sub Deconstruct!" << std::endl;
	}

};

int main()
{
	BaseClass *p_b = new BaseClass;
	delete p_b;
	std::cout << "---------------------------------\n";
	SubClass *p_s = new SubClass;
	delete p_s;
	std::cout << "---------------------------------\n";
	BaseClass *p_b_s = new SubClass;
	delete p_b_s;

	std::cin.get();
}


从如上例子可以看出子类的析构函数未被调用,此时造成内存泄露。但是在基类加上virtual后,如下


#include<iostream>

class BaseClass
{
private:
	int *m_Basepointer;
public:
	BaseClass() { std::cout << "it's Base Construct!" << std::endl; }
    virtual	~BaseClass()
	{
		delete m_Basepointer;
		std::cout << "it's Base Deconstruct!" << std::endl;
	}
};

class SubClass:public BaseClass
{
public:
	SubClass() { std::cout << "it's Sub Construct!" << std::endl; }
	~SubClass() 
	{
		std::cout << "it's Sub Deconstruct!" << std::endl;
	}

};

int main()
{
	BaseClass *p_b = new BaseClass;
	delete p_b;
	std::cout << "---------------------------------\n";
	SubClass *p_s = new SubClass;
	delete p_s;
	std::cout << "---------------------------------\n";
	BaseClass *p_b_s = new SubClass;
	delete p_b_s;

	std::cin.get();
}

成功释放,这就是为何要在基类加上virtual防止内存泄露,保证安全释放。

posted @ 2020-03-18 12:03  Yekko  阅读(149)  评论(0编辑  收藏  举报