juc · 2019-09-19 0

CountDownLatch、CyclicBarrier、Semaphore的使用

一、CountDownLatch

CountDownLatch(倒计时计数器、闭锁),让一些线程阻塞直到另一些线程完成一些操作才被唤醒

CountDownLatch主要有两个方法:当一个或多个线程调用await()方法时,调用线程会被阻塞;其他线程调用countDown()方法会将计数器减一,调用countDown()方法不会被阻塞

当计数器的值为0时,调用await()方法被阻塞的线程会被唤醒

public class CountDownLatchTest {

    public static void main(String[] args) throws InterruptedException {
        CountDownLatch countDownLatch = new CountDownLatch(5);
        for(int i = 0; i < 5; i++){
            new Thread(()->{
                try {
                    TimeUnit.SECONDS.sleep(1);//睡眠1秒
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName() + "t 线程");
                countDownLatch.countDown(); //计数器减一
            }, String.valueOf(i)).start();
        }

        countDownLatch.await();//计数器为0时,唤醒线程
        System.out.println(Thread.currentThread().getName() + "t 线程");
    }

}

结果:

CountDownLatch

二、CyclicBarrier

CyclicBarrier(可循环屏障,栅栏),让一组线程到达一个屏障(同步点)时被阻塞,直到最后一个线程到达屏障时,屏障才会开门,线程才会继续执行

public class CyclicBarrierTest {
    public static void main(String[] args) {
//        CyclicBarrier cyclicBarrier = new CyclicBarrier(3);
        CyclicBarrier cyclicBarrier = new CyclicBarrier(3, ()->{
            System.out.println(Thread.currentThread().getName() + "t work");
        });
        for(int i = 0; i < 6; i++){
            new Thread(()->{
                System.out.println(Thread.currentThread().getName() + "t begin");
                try {
                    cyclicBarrier.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                } catch (BrokenBarrierException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName() + "t end");
            }, String.valueOf(i)).start();
        }

    }
}

结果:

CyclicBarrier

三、Semaphore

Semaphore(信号量),主要用于两个目的,一个是用于多个共享资源的互斥作用,另一个用于并发线程数的控制

public class SemaphoreDemo {
    public static void main(String[] args) {
        Semaphore semaphore = new Semaphore(3);//模拟3个停车位
        for(int i = 0; i < 6; i++){//模拟6部汽车
            new Thread(()->{
                try {
                    semaphore.acquire();
                    int time = (int)(1+Math.random()*(10-1+1));
                    System.out.println(Thread.currentThread().getName() + "t抢到车位, 使用"+time+"秒");
                    TimeUnit.SECONDS.sleep(time);
                    System.out.println(Thread.currentThread().getName() + "t离开车位");
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }finally {
                    semaphore.release();
                }
            }, String.valueOf(i)).start();
        }

    }
}

结果:

Semaphore