1.思维导图
2.作业:
编程题:
以下是一个简单的比喻,将多态概念与生活中的实际情况相联系:
比喻:动物园的讲解员和动物表演
想象一下你去了一家动物园,看到了许多不同种类的动物,如狮子、大象、猴子等。现在,动物园里有一位讲解员,他会为每种动物表演做简单的介绍。
在这个场景中,我们可以将动物比作是不同的类,而每种动物表演则是类中的函数。而讲解员则是一个基类,他可以根据每种动物的特点和表演,进行相应的介绍。
具体过程如下:
定义一个基类 Animal,其中有一个虛函数perform(),用于在子类中实现不同的表演行为。
#include <iostream>
using namespace std;
class Animal
{
public:
string name;
int weight;
string color;
string sex;
public:
Animal(){}
Animal(string n,int w,string c,string s):name(n),weight(w),color(c),sex(s)
{
cout << "我是动物园的讲解员,下面由我来给大家介绍这里的东动物"<< endl;
}
//构造纯虚函数
virtual void perfrom()
{
}
void show()
{
cout << "name: "<< name<<endl;
cout << "weight:" <<weight<< endl;
cout << "color:"<< color<<endl;
cout << "sex:"<< sex<<endl;
}
};
class Lion:public Animal
{
private:
string say;
public:
Lion(){}
Lion (string n ,int w,string c,string s,string a):Animal(n,w,c,s),say(a)
{
}
void perfrom()
{
cout << "Lion:我要表演狮子怒吼" << endl;
}
void show()
{
cout << "name: "<< name<<endl;
cout << "weight:" <<weight<< endl;
cout << "color:"<< color<<endl;
cout << "sex:"<< sex<<endl;
cout << "技能:say:"<< say <<endl;
}
};
class Elephant:public Animal
{
private:
string water;
public:
Elephant(){}
Elephant(string n,int w,string c,string s,string wa):Animal(n,w,c,s),water(wa)
{
}
void perfrom()
{
cout << "elephant:我要表演大象喷水" << endl;
}
void show()
{
cout << "name: "<< name<<endl;
cout << "weight:" <<weight<< endl;
cout << "color:"<< color<<endl;
cout << "sex:"<< sex<<endl;
cout << "技能:water:"<< water <<endl;
}
};
class Monkey:public Animal
{
private:
string eat;
public:
Monkey(){}
Monkey(string n,int w,string c,string s,string e):Animal(n,w,c,s),eat(e)
{
}
void perfrom()
{
cout << "monkey: 我要表演猴子吃桃" << endl;
}
void show()
{
cout << "name: "<< name<<endl;
cout << "weight:" <<weight<< endl;
cout << "color:"<< color<<endl;
cout << "sex:"<< sex<<endl;
cout << "技能:eat:"<< eat <<endl;
}
};
int main()
{
Animal *p;
Lion s("狮子",200,"yellow","公","表演节目:狮子怒吼");
Elephant e("大象",20000,"blue","母","节目:大象喝水");
Monkey m("猴子",10,"bleak","公","节目:猴子吃桃");
s.show();
p=&s;
p->perfrom();
e.show();
p=&e;
p->perfrom();
m.show();
p=&m;
p->perfrom();
return 0;
}
效果图: