5

根据 Javadoc： CountDownLatch 使用给定的计数进行初始化。await 方法阻塞，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();
        }
    }

}
4

2 回答 2

6

尝试运行以下修改后的代码版本：

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latch.await()在第一个线程调用之后从调用中解除阻塞latch.countDown()

于 2012-03-23T04:42:14.793 回答
1

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查看它按您的预期工作。

于 2012-03-23T03:07:06.063 回答