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技术 2022年11月18日
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介绍

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

实现原理

计数器通过使用锁(共享锁、排它锁)实现

实例1

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.

注:countDown()一定要执行到(考虑异常及线程与开始计数设置不一致),否则会一直卡在await()(可以设置时间,超过一定时间就不等了)

实例2(和join的相似处)

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

实例3(和join的不同处)

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
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