多线程同步中的wait,notify,notifyAll方法的深入学习

来源:互联网 发布:九搜网络用户中心 编辑:程序博客网 时间:2024/05/18 01:50

在多线程的运用中,难免希望某个线程暂停或唤醒某个线程,比如希望操作不同功能的线程交替执行。因此,java提供了wait,notify,和notifyAll方法去实现该效果。

wait():用于冻结执行该语句的线程,使其失去执行权,wait方法返回值是异常,所以要try,线程被冻结后,同步对其失效,也就是说该线程在同步代码中冻结了,其他线程可以进入同步代码。

notify():用于唤醒所有处于被冻结状态的线程中最早被冻结的线程。

notifyAll():用于唤醒所有处于冻结状态的线程。

要注意的是,调用方法是通过“锁”对象去点调用这些方法,而且作用的线程局限于被该“锁“控制的线程,而不是程序中所有线程。

实例解释:

//需求:产品每生产一个消费一个class ProducerConsumerDemo {public static void main(String[] args) {Resource r = new Resource();Producer pro = new Producer(r);Consumer con = new Consumer(r);Thread t1 = new Thread(pro);Thread t2 = new Thread(pro);Thread t3 = new Thread(con);Thread t4 = new Thread(con);t1.start();t2.start();t3.start();t4.start();}}class Resource{private String name;private int count = 1;private boolean flag = false;public synchronized void set(String name){if(flag)try{this.wait();}catch(Exception e){}this.name = name+"--"+count++;System.out.println(Thread.currentThread().getName()+"...生产者.."+this.name);flag = true;this.notify();}public synchronized void out(){if(!flag)try{wait();}catch(Exception e){}System.out.println(Thread.currentThread().getName()+"...消费者........."+this.name);flag = false;this.notify();}}class Producer implements Runnable{private Resource res;Producer(Resource res){this.res = res;}public void run(){while(true){res.set("+商品+");}}}class Consumer implements Runnable{private Resource res;Consumer(Resource res){this.res = res;}public void run(){while(true){res.out();}}}
但是结果却不尽如人意:


如图所示,同一商品被卖了两次,这是为什么呢?

原因是,在执行out方法时,如果有一个线程被冻结了,那么当它被唤醒时,它不会再判断flag,而直接执行后面语句,从而导致该错误,同样,改程序有可能出现生产者生产两个产品去没人买。

要想解决该问题,只能让线程被唤醒后,要重新判断flag,因此把if语句改成while

//需求:产品每生产一个消费一个class ProducerConsumerDemo {public static void main(String[] args) {Resource r = new Resource();Producer pro = new Producer(r);Consumer con = new Consumer(r);Thread t1 = new Thread(pro);Thread t2 = new Thread(pro);Thread t3 = new Thread(con);Thread t4 = new Thread(con);t1.start();t2.start();t3.start();t4.start();}}class Resource{private String name;private int count = 1;private boolean flag = false;public synchronized void set(String name){while(flag)try{this.wait();}catch(Exception e){}this.name = name+"--"+count++;System.out.println(Thread.currentThread().getName()+"...生产者.."+this.name);flag = true;this.notify();}public synchronized void out(){while(!flag)try{wait();}catch(Exception e){}System.out.println(Thread.currentThread().getName()+"...消费者........."+this.name);flag = false;this.notify();}}class Producer implements Runnable{private Resource res;Producer(Resource res){this.res = res;}public void run(){while(true){res.set("+商品+");}}}class Consumer implements Runnable{private Resource res;Consumer(Resource res){this.res = res;}public void run(){while(true){res.out();}}}
结果很遗憾,比上次更惨,中途直接停住了:


这是因为,改为while后,可能会出现线程全都冻结的状况。。。。(t1执行set,再执行一次就冻结了,t2执行set,冻结了,t3执行out,t1解冻,t3再执行,冻结了,t4执行out,冻结了,t1执行set,t2解冻,t1再执行set,冻结了,t2执行set,冻结了,这时全部冻结)

为了排除全部冻结的情况,应该把notify改为notifyAll

//需求:产品每生产一个消费一个class ProducerConsumerDemo {public static void main(String[] args) {Resource r = new Resource();Producer pro = new Producer(r);Consumer con = new Consumer(r);Thread t1 = new Thread(pro);Thread t2 = new Thread(pro);Thread t3 = new Thread(con);Thread t4 = new Thread(con);t1.start();t2.start();t3.start();t4.start();}}class Resource{private String name;private int count = 1;private boolean flag = false;public synchronized void set(String name){while(flag)try{this.wait();}catch(Exception e){}this.name = name+"--"+count++;System.out.println(Thread.currentThread().getName()+"...生产者.."+this.name);flag = true;this.notifyAll();}public synchronized void out(){while(!flag)try{wait();}catch(Exception e){}System.out.println(Thread.currentThread().getName()+"...消费者........."+this.name);flag = false;this.notifyAll();}}class Producer implements Runnable{private Resource res;Producer(Resource res){this.res = res;}public void run(){while(true){res.set("+商品+");}}}class Consumer implements Runnable{private Resource res;Consumer(Resource res){this.res = res;}public void run(){while(true){res.out();}}}
结果正确!!

文章中实例来自网络



0 0
原创粉丝点击