关于继承和虚函数的入门讨论

最新推荐文章于 2025-09-09 16:41:33 发布

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文章标签：面试

原文链接：http://www.cnblogs.com/lihaozy/archive/2010/11/01/1865898.html

本文通过一个C++示例程序，详细解析了虚函数和非虚函数在继承类中的调用机制。演示了当使用基类指针指向派生类对象时，虚函数如何实现动态绑定。

前几天去面试，碰到下面一道题目，问输出是什么？

#include <iostream>
using namespace std;

class Base
{
public:
	virtual void function()
	{
		cout<<"function in Base\n";
	}

	void function (int param)
	{
		cout<<"function in Base with param\n";
	}
};

class Derived : public Base
{
public:
	virtual void function()
	{
		cout<<"function in Derived\n";
	}
	void function(int param)
	{
		cout<<"function in Derived with param\n";
	}
};

void main()
{
	Base base;
	Derived derived;
	
	base.function();
	base.function(1);
	derived.function();
	derived.function(1);

	Base * print = (Base *)&derived;
	print->function();
	print->function(1);
}

以下有solution的part：

#include <iostream>
using namespace std;

class Base
{
public:
	virtual void function()
	{
		cout<<"function in Base\n";
	}

	void function (int param)
	{
		cout<<"function in Base with param\n";
	}
};

class Derived : public Base
{
public:
	virtual void function()
	{
		cout<<"function in Derived\n";
	}
	void function(int param)
	{
		cout<<"function in Derived with param\n";
	}
};

void main()
{
	Base base;
	Derived derived;
	
	base.function();//function in Base，没有疑问
	base.function(1);//function in Base with param，没有疑问
	derived.function();//function in Derived，没有疑问
	derived.function(1);//function in Derived with param，没有疑问

	Base * print = (Base *)&derived;//这里加不加(Base *)，下面的结果都是相同的
	print->function();//这里，虽然是Base类型的指针，但是因为是虚函数的原因，有vbtl做指示，所以总是可以判断出属于哪一层
	print->function(1);//在这里，既然是Base类型的指针，那么其function()就会调用Base中的function，function in Base with param
}

转载于:https://www.cnblogs.com/lihaozy/archive/2010/11/01/1865898.html