• 面向对象综合练习题(动物乐园)


    动物乐园

    任务描述:

      以面向对象思想设计动物乐园系统

      动物乐园包括的成员:猫、鸭子,海豚,……

      属性:名称、腿

      方法:叫

      设计类图

      使用多态、接口优化设计

     

      编码实现

    分别创建Cat、Duck、Dolphin对象并放到一个数组中,对数组进行遍历输出各种动物如何叫,各种动物腿的条数。instanceof判断动物种类     

     

    package com.Summer_0428.cn;
    
    abstract class Animal1{
        private String name;
        public Animal1(String name) {
            this.name = name;
        }
        
        public String getName() {
            return name;
        }
        public abstract void show();//定义一个抽象方法
    }
    
    interface Terrestrial{
        public int getLegNum();
    }
    
    class Cat1 extends Animal1 implements Terrestrial{
        private int legNum;
        public Cat1(String name,int legNum) {
            super(name);
            this.legNum = legNum;
        }
    
        @Override
        public int getLegNum() {
            return legNum;
        }
        
        @Override
        public void show() {
            System.out.println("喵喵喵");
        }
        
    }
    
    class Duck extends Animal1 implements Terrestrial{
        private int legNum;
        public Duck(String name,int legNum) {
            super(name);
            this.legNum = legNum;
        }
    
        @Override
        public int getLegNum() {
            return legNum;
        }
        
        @Override
        public void show() {
            System.out.println("嘎嘎嘎");
        }
        
    }
    
    class Dolphin extends Animal1 {
    
        public Dolphin(String name) {
            super(name);
        }
        
        @Override
        public void show() {
            System.out.println("海豚音");
        }
        
    }
    public class Zoo {
    
        public static void main(String[] args) {
            Animal1 animals[] = new Animal1[3];
            animals[0] = new Cat1("加菲猫", 4);
            animals[1] = new Duck("唐小鸭", 2);
            animals[2] = new Dolphin("海豚奇奇");
            
            System.out.println("动物名字		腿的条数		动物叫");
            for (int i = 0; i < animals.length; i++){
                if (animals[i] instanceof Terrestrial) {
                    String name = animals[i].getName();
                    int legNum = ((Terrestrial) animals[i]).getLegNum();
                    System.out.print(name + "		" + legNum + "		");
                    animals[i].show();
                } else {
                    String name = animals[i].getName();
                    System.out.print(name + "		" + 0 + "		");
                    animals[i].show();
                }
            }
            
    
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/summerdata/p/10787791.html
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