java多线程问题集锦(一)

问题:设计4个线程，其中两个线程每次对j增加1，另外两个线程对j每次减少1，运行1000次(可以每个各执行250次也可自己计算只要总合1000)，J初始为100，保证结果为100，写出程序

PS：这里主要用到了synchronized以及对象的wait和notify以及notifyAll方法

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package com.test;public class TestThread{public static void main(String[] args) {Operate o = new Operate();IncreaseThread thread1 = new IncreaseThread(o);IncreaseThread thread2 = new IncreaseThread(o);DecreaseThread thread3 = new DecreaseThread(o);DecreaseThread thread4 = new DecreaseThread(o);thread1.start();thread2.start();thread3.start();thread4.start();}}class Operate{private int j=100;private int count = 0;public synchronized void increase(){System.out.println("increase in");while(100!=j){try {System.out.println(Thread.currentThread().getName()+"increase begin wait");wait();System.out.println(Thread.currentThread().getName()+"increase end wait");} catch (InterruptedException e) {e.printStackTrace();}}j++;System.out.println("increase j:"+j);count++;System.out.println("increase count:"+count);//切勿使用notify();只唤醒一个线程是不行的，因为有两个相同功能的线程会导致这2个线程都wait然后再通知到另外2个相同功能的线程又会2个线程都wait最终可能全部4个线程全部waitnotifyAll();System.out.println("increase out");}public synchronized void decrease(){System.out.println("decrease in");while(100==j){try {System.out.println(Thread.currentThread().getName()+"decrease begin wait");wait();System.out.println(Thread.currentThread().getName()+"decrease end wait");} catch (InterruptedException e) {e.printStackTrace();}}j--;System.out.println("decrease j:"+j);count++;System.out.println("decrease count:"+count);//切勿使用notify();只唤醒一个线程是不行的，因为有两个相同功能的线程会导致这2个线程都wait然后再通知到另外2个相同功能的线程又会2个线程都wait最终可能全部4个线程全部waitnotifyAll();System.out.println("decrease out");}}class IncreaseThread extends Thread{private  Operate o;public IncreaseThread(Operate o) {this.o = o;}@Overridepublic void run() {for(int i = 0;i<250;i++){o.increase();}}}class DecreaseThread extends Thread{private  Operate o;public DecreaseThread(Operate o) {this.o = o;}@Overridepublic void run() {for(int i = 0;i<250;i++){o.decrease();}}}