标签：ext interrupt 不执行 count() tar BMI test 多个 ted
CountDownLatch是一个同步辅助类，它允许一个或多个线程一直等待直到其他线程执行完毕才开始执行。
用给定的计数初始化CountDownLatch，其含义是要被等待执行完的线程个数。
每次调用CountDown()，计数减1
主程序执行到await()函数会阻塞等待线程的执行，直到计数为0
计数器通过使用锁（共享锁、排它锁）实现
场景：模拟10人赛跑。10人跑完后才喊"Game Over."
package com.jihite; import java.util.concurrent.CountDownLatch; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; public class CountDownLatchTest { private static final int RUNNER_COUNT = 10; public static void main(String[] args) throws InterruptedException { final CountDownLatch begin = new CountDownLatch(1); final CountDownLatch end = new CountDownLatch(RUNNER_COUNT); final ExecutorService exec = Executors.newFixedThreadPool(10); for (int i = 0; i < RUNNER_COUNT; i++) { final int NO = i + 1; Runnable run = new Runnable() { @Override public void run() { try { begin.await(); Thread.sleep((long)(Math.random() * 10000)); System.out.println("No." + NO + " arrived"); } catch (InterruptedException e) { e.printStackTrace(); } finally { end.countDown(); } } }; exec.submit(run); } System.out.println("Game Start ..."); begin.countDown(); end.await(); // end.await(30, TimeUnit.SECONDS); System.out.println("Game Over."); exec.shutdown(); } }
分析：代码中定义了2个计数器，个数分别为1和10。
如果不执行begin.countDown()，进程会一致阻塞在begin.await()
主进程执行到end.awit()阻塞等待end计数器清0，进程中每执行一次CountDown()减1，所有执行完后主进程继续往下执行
输出
Game Start ... No.6 arrived No.4 arrived No.10 arrived No.3 arrived No.9 arrived No.5 arrived No.8 arrived No.7 arrived No.1 arrived No.2 arrived Game Over.
场景：流水线上有3个worker: worker1、worker2、worker3,只有当worker1和worker2执行完时才可以执行worker3
WorkerCount.java
package com.jihite; import java.util.concurrent.CountDownLatch; public class WorkerCount extends Thread { private String name; private long time; private CountDownLatch countDownLatch; public WorkerCount(String name, long time, CountDownLatch countDownLatch) { this.name = name; this.time = time; this.countDownLatch = countDownLatch; } @Override public void run() { try { System.out.println(name + "开始工作"); Thread.sleep(time); System.out.println(name + "工作完成, 耗时："+ time); countDownLatch.countDown(); System.out.println("countDownLatch.getCount():" + countDownLatch.getCount()); } catch (InterruptedException e) { e.printStackTrace(); } } }
CountDownLatch实现：
@Test public void CountDownLatchTest() throws InterruptedException { int COUNT = 2; final CountDownLatch countDownLatch = new CountDownLatch(COUNT); WorkerCount worker0 = new WorkerCount("lilei-0", (long)(Math.random() * 10000), countDownLatch); WorkerCount worker1 = new WorkerCount("lilei-1", (long)(Math.random() * 10000), countDownLatch); worker0.start(); worker1.start(); countDownLatch.await(); System.out.println("准备工作就绪"); WorkerCount worker2 = new WorkerCount("lilei-2", (long)(Math.random() * 10000), countDownLatch); worker2.start(); Thread.sleep(10000); }
输出：
lilei-0开始工作 lilei-1开始工作 lilei-1工作完成, 耗时：4039 countDownLatch.getCount():1 lilei-0工作完成, 耗时：9933 countDownLatch.getCount():0 准备工作就绪 lilei-2开始工作 lilei-2工作完成, 耗时：6402 countDownLatch.getCount():0
该场景join也可以完成
Worker.java
package com.jihite; public class Worker extends Thread{ private String name; private long time; public Worker(String name, long time) { this.name = name; this.time = time; } @Override public void run() { try { System.out.println(name + "开始工作"); Thread.sleep(time); System.out.println(name + "工作完成, 耗时："+ time); } catch (InterruptedException e) { e.printStackTrace(); } } }
join实现
@Test public void JoinTest() throws InterruptedException { Worker worker0 = new Worker("lilei-0", (long)(Math.random() * 10000)); Worker worker1 = new Worker("lilei-1", (long)(Math.random() * 10000)); Worker worker2 = new Worker("lilei-2", (long)(Math.random() * 10000)); worker0.start(); worker1.start(); worker0.join(); worker1.join(); System.out.println("准备工作就绪"); worker2.start(); Thread.sleep(10000); }
输出
lilei-0开始工作 lilei-1开始工作 lilei-1工作完成, 耗时：4483 lilei-0工作完成, 耗时：6301 准备工作就绪 lilei-2开始工作 lilei-2工作完成, 耗时：6126
既然这样，那CountDownLatch和join的区别在哪？通过下面的场景三就可以看出
场景：流水线上有3个worker: worker1、worker2、worker3,只有当worker1和worker2两者的阶段一都执行完后才可以执行worker3
WorkerCount2.java
package com.jihite; import java.util.concurrent.CountDownLatch; public class WorkerCount2 extends Thread { private String name; private long time; private CountDownLatch countDownLatch; public WorkerCount2(String name, long time, CountDownLatch countDownLatch) { this.name = name; this.time = time; this.countDownLatch = countDownLatch; } @Override public void run() { try { System.out.println(name + "开始阶段1工作"); Thread.sleep(time); System.out.println(name + "阶段1完成, 耗时："+ time); countDownLatch.countDown(); System.out.println(name + "开始阶段2工作"); Thread.sleep(time); System.out.println(name + "阶段2完成, 耗时："+ time); } catch (InterruptedException e) { e.printStackTrace(); } } }
此时用join无法实现，只能用CountDownLatch
@Test public void CountDownLatchTest2() throws InterruptedException { int COUNT = 2; final CountDownLatch countDownLatch = new CountDownLatch(COUNT); WorkerCount2 worker0 = new WorkerCount2("lilei-0", (long)(Math.random() * 10000), countDownLatch); WorkerCount2 worker1 = new WorkerCount2("lilei-1", (long)(Math.random() * 10000), countDownLatch); worker0.start(); worker1.start(); countDownLatch.await(); System.out.println("准备工作就绪"); WorkerCount2 worker2 = new WorkerCount2("lilei-2", (long)(Math.random() * 10000), countDownLatch); worker2.start(); Thread.sleep(10000); }
输出
lilei-0开始阶段1工作 lilei-1开始阶段1工作 lilei-0阶段1完成, 耗时：3938 lilei-0开始阶段2工作 lilei-1阶段1完成, 耗时：6259 lilei-1开始阶段2工作 准备工作就绪 lilei-2开始阶段1工作 lilei-0阶段2完成, 耗时：3938 lilei-1阶段2完成, 耗时：6259 lilei-2阶段1完成, 耗时：7775 lilei-2开始阶段2工作
标签：ext interrupt 不执行 count() tar BMI test 多个 ted
原文地址：https://www.cnblogs.com/kaituorensheng/p/9043494.html
