Java第十次作业

(一)学习总结

1.用思维导图对java多线程的学习内容进行总结

2.下面是一个单线程实现的龟兔赛跑游戏。

public class TortoiseHareRace {
    public static void main(String[] args) {
        int totalStep = 10;
        int tortoiseStep = 0;
        int hareStep = 0;
        boolean[] flags = {true,false};
        System.out.println("龟兔赛跑开始了...");
        while(tortoiseStep < totalStep && hareStep < totalStep){
            tortoiseStep++;
            System.out.println("乌龟跑了"+tortoiseStep+"步...");
            boolean isHareSleep = flags[((int)(Math.random()*10))%2];
            if(isHareSleep){
                System.out.println("兔子睡着了zzzz");
            }else{
                hareStep += 2;
                System.out.println("兔子跑了"+hareStep+"步...");
            }
        }       
    }
}

阅读程序,采用实现Runnable接口的方式用多线程实现这个小游戏。下面给出主线程类,补充Tortoise线程类和Hare线程类。

public class TortoiseHareRace { 
    public static void main(String[] args) {
        Tortoise tortoise = new Tortoise(10);
        Hare hare = new Hare(10);
        Thread tortoiseThread = new Thread(tortoise);
        Thread hareThread = new Thread(hare);
        tortoiseThread.start();
        hareThread.start();
    }
}

补充:

public class TortoiseHareRace { 
    public static void main(String[] args) {
        Tortoise tortoise = new Tortoise(10);
        Hare hare = new Hare(10);
        Thread tortoiseThread = new Thread(tortoise);
        Thread hareThread = new Thread(hare);
        tortoiseThread.start();
        hareThread.start();
    }
}

class Hare implements Runnable{
    int hareStep=0;
    int totalStep=0;
    boolean[] flags = {true,false};
    boolean isHareSleep = flags[((int)(Math.random()*10))%2];
    public Hare(int totalStep){
        this.totalStep=totalStep;
    }
    public void run(){
        for(hareStep=0;hareStep<10;hareStep+=2){
            if(isHareSleep){
                System.out.println("兔子睡着了zzzz");
            }else{
                System.out.println("兔子跑了"+(hareStep+1)+"步...");
            }
        }
    }
}

class Tortoise implements Runnable{
    int tortoiseStep=0;
    int totalStep=0;
    public Tortoise(int totalStep){
        this.totalStep=totalStep;
    }
    public void run(){
        for(tortoiseStep=0;tortoiseStep<10;tortoiseStep++){
            System.out.println("乌龟跑了"+(tortoiseStep+1)+"步...");
        }
    }
}

3.下面的程序是模拟了生产者——消费者问题,生产者生产10个数,消费者依次消费10个数,运行程序,看结果是否正常?存在什么问题?说明原因。使用synchronized, wait, notify解决程序出现的问题。写出修改的部分程序即可。

class Consumer implements Runnable {
    private Clerk clerk;
    public Consumer(Clerk clerk) {
        this.clerk = clerk;
    }
    public void run() {
        System.out.println("消费者开始消耗整数......");
        // 消耗10个整数
        for(int i = 1; i <= 10; i++) {
            try {
                 // 等待随机时间
                Thread.sleep((int) (Math.random() * 3000));
            }
            catch(InterruptedException e) {
                e.printStackTrace();
            }              
            clerk.getProduct();// 从店员处取走整数
        }
    }
 }
class Producer implements Runnable {
    private Clerk clerk;
    public Producer(Clerk clerk) {
        this.clerk = clerk;
    }
    public void run() {
        System.out.println( "生产者开始生产整数......");
        // 生产1到10的整数
        for(int product = 1; product <= 10; product++) {
            try {
                Thread.sleep((int) Math.random() * 3000);
            }
            catch(InterruptedException e) {
                e.printStackTrace();
            }
           clerk.setProduct(product); // 将产品交给店员
        }
    } 
}
public class ProductTest {
    public static void main(String[] args) {
        Clerk clerk = new Clerk();
        Thread consumerThread = new Thread(new Consumer(clerk)); 
        Thread producerThread = new Thread(new Producer(clerk)); 
        consumerThread.start(); 
        producerThread.start(); 
    }
}
class Clerk {
    private int product = -1; // -1 表示目前没有产品 
     // 这个方法由生产者呼叫
    public void setProduct(int product) {
        this.product = product; 
        System.out.printf("生产者设定 (%d)%n", this.product);      
    } 
    // 这个方法由消费者呼叫
    public int getProduct() {          
        int p = this.product; 
        System.out.printf("消费者取走 (%d)%n", this.product);      
        return p; 
    } 
}

修改:

class Clerk {// 定义信息类
    private int clerk;
    private boolean flag = true; // 起始状态先生产后消费

    
    public synchronized void setProduct(int clerk) {
        // flag = false 无法生成
        while (this.flag == false) {
            
            try {
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        this.clerk = clerk;
        System.out.printf("生产者设定 (%d)%n", this.clerk); 
        this.flag = false;
        notify();
    }

    public synchronized int getProduct() {
        // flag = true 无法取走
         (this.flag == true) {
            try {
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        this.flag = true;
        notify();
        System.out.printf("消费者取走 (%d)%n",this.clerk);
        return clerk;
    }
}

(二)实验总结

实验内容:
1.模拟三个老师同时分发80分作业,每个老师相当于一个线程。

2.模拟一个银行存款的程序。假设有两个储户都去银行往同一个账户进行存款,一次存100,每人存三次。要求储户每存一次钱,账户余额增加100,并在控制台输出当前账户的余额。

(三)代码托管
https://git.oschina.net/hebau_cs15/Java-CS01QH.git

原文地址:https://www.cnblogs.com/JoeH/p/6937497.html