关于继承和虚函数的入门讨论

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
}