CountDownLatch - 理解 await 和 countDown

Question

根据 Javadoc： CountDownLatch 使用给定的计数进行初始化。 await 方法会一直阻塞，直到当前计数达到零。

这意味着在下面的代码中，因为我将 CountDownLatch 初始化为 1。一旦锁存器调用倒计时，所有线程都应该从它的 await 方法中解除阻塞。

但是，主线程正在等待所有线程完成。而且，我没有将主线程加入到其他线程的末尾。为什么主线程在等待？

    import java.util.concurrent.CountDownLatch;
    import java.util.concurrent.atomic.AtomicLong;

public class Sample implements Runnable {

    private CountDownLatch latch;

    public Sample(CountDownLatch latch)
    {
        this.latch = latch;
    }
    private static AtomicLong number = new AtomicLong(0);

    public long next() {
         return number.getAndIncrement();
    }

    public static void main(String[] args) {

        CountDownLatch latch = new CountDownLatch(1);
        for (int threadNo = 0; threadNo < 4000; threadNo++) {
          Runnable t = new Sample(latch);
          new Thread(t).start();
        }
        try {
            latch.countDown();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    @Override
    public void run() {
        try {
            latch.await();
            Thread.sleep(100);
            System.out.println("Count:"+next());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

}

Answer 1

尝试运行您的代码的以下修改版本：

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.atomic.AtomicLong;

public class Test implements Runnable {

    private CountDownLatch latch;

    public Test(CountDownLatch latch)
    {
        this.latch = latch;
    }
    private static AtomicLong number = new AtomicLong(0);

    public long next() {
         return number.getAndIncrement();
    }

    public static void main(String[] args) {

        CountDownLatch latch = new CountDownLatch(1);
        for (int threadNo = 0; threadNo < 1000; threadNo++) {
          Runnable t = new Test(latch);
          new Thread(t).start();
        }
        try {
            latch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }

        System.out.println( "done" );
    }

    @Override
    public void run() {
        try {
            Thread.sleep(1000 + (int) ( Math.random() * 3000 ));
            System.out.println(next());
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            latch.countDown();
        }
    }

}

您应该会看到如下内容：

0
完成
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

这表明主线程实际上在第一个线程调用latch.countDown() 之后解除了对latch.await() 调用的阻塞

Answer 2

您正在启动 4000 个线程，它们只等待 100 毫秒。你很可能压倒了盒子（所有线程将在大致相同的时间结束）。在您的线程开始外观中添加一个睡眠，并尝试增加超时以使其按预期工作。