Java第十次作业--多线程
(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();
}
}
- 补充后:
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();
}
}
public class Tortoise implements Runnable {
int totalStep = 10;
int tortoiseStep = 0;
public void run() {
while (tortoiseStep < totalStep) {
synchronized (this) {
tortoiseStep++;
System.out.println("乌龟跑了" + tortoiseStep + "步...");
}
}
}
}
public class Hare implements Runnable {
int totalStep = 10;
int hareStep = 0;
boolean[] flags = { true, false };
public void run() {
while (hareStep < totalStep) {
synchronized (this) {
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...");
}
}
}
}
}
public class Test {
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();
}
}
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;
}
}
- 运行结果为:
- 代码修改:
class Clerk {// 定义信息类
private int clerk;
private boolean flag = true; // 起始状态先生产后消费
// flag = true 可以生产不能取走 flag = false 可以取走不能生成
public synchronized void setProduct(int clerk) {
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() {
while (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
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。
- 程序设计思路:
创建一个Teacher类继承Thread;将线程中要执行的代码写在run()方法中,覆盖父类中的run();创建测试类,调用start()方法启动线程,自动运行run()方法。
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。
- 程序设计思路:
银行类:定义一个存储账户余额的变量以及一个存款的方法
储户类:实现向账户存款3次。每一个储户是一个线程
测试类:创建客户对象,启动线程