本文共 2738 字，大约阅读时间需要 9 分钟。

1.CountDownLatch

package com.company.common_assist;public class CountDownLatchTest {       public static void main(String[] args) throws InterruptedException {           //总数是6，必须要执行任务的时候，再使用        java.util.concurrent.CountDownLatch countDownLatch =  new java.util.concurrent.CountDownLatch(5);        for (int i = 1; i <= 6; i++) {               new Thread(()->{                   System.out.println(Thread.currentThread().getName() + "go out");                try {                       countDownLatch.await();                } catch (InterruptedException e) {                       e.printStackTrace();                }            },String.valueOf(i)).start();        }        countDownLatch.await(); //等待计数器归零，然后向下执行        System.out.println("Close Door");    }}

运行结果：

2.CyclicBarrier

package com.company.common_assist;import java.util.concurrent.BrokenBarrierException;import java.util.concurrent.CyclicBarrier;//允许一组线程全部等待彼此到达共用的同步复制public class CyclicBarrierTest {       public static void main(String[] args) {           CyclicBarrier cyclicBarrier = new CyclicBarrier(7,()->{               System.out.println("召唤神龙");        });        for (int i = 1; i <= 7; i++) {               final int temp = i;            new Thread(()->{                   System.out.println(Thread.currentThread().getName() + "收集第" + temp + "个龙珠");                try {                       cyclicBarrier.await(); //等待                } catch (InterruptedException e) {                       e.printStackTrace();                } catch (BrokenBarrierException e) {                       e.printStackTrace();                }            },String.valueOf(i)).start();        }    }}

运行结果：

3.Semaphore

package com.company.common_assist;import java.util.concurrent.Semaphore;import java.util.concurrent.TimeUnit;//一个计数信号量，在概念上，信号量要维持一组许可证//如果有必要，每个acquire()都会阻塞，直到许可证可用，然后才能使用它//作用：多个共享资源互斥的使用！并发限流，控制最大的线程数public class SemaphoreTest {       public static void main(String[] args) {           //线程数量：停车位！限流！        Semaphore semaphore = new Semaphore(3);        for (int i = 1; i <= 6; i++) {               new Thread(()->{                   try {                       //如果已经满了，会等待资源被释放                    semaphore.acquire();                    System.out.println(Thread.currentThread().getName() + "抢到车位");                    TimeUnit.SECONDS.sleep(2);                    System.out.println(Thread.currentThread().getName() + "离开车位");                } catch (InterruptedException e) {                       e.printStackTrace();                }finally {                       semaphore.release(); //释放，然后唤醒等待的资源                }            },String.valueOf(i)).start();        }    }}

运行结果：

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