Question

我已经开发了一个样本Java程序来了解countdownlatch并使用count 4初始化了countdownlatch。我希望在countDown方法之后，getCount（）将返回countdownlatch的剩余计数。但是，在以下示例中：-

public static void main(String args[]) throws InterruptedException {
    CountDownLatch latch = new CountDownLatch(4);
    Worker first = new Worker(latch, "WORKER-1");
    Worker second = new Worker(latch, "WORKER-2");
    Worker third = new Worker(latch, "WORKER-3");
    Worker fourth = new Worker(latch, "WORKER-4");
    first.start();
    second.start();
    third.start();
    fourth.start();

    latch.await();
    System.out.println("Final Count:- " + latch.getCount());
}
}


class Worker extends Thread {
private CountDownLatch latch;

public Worker(CountDownLatch latch, String name) {
    super(name);
    this.latch = latch;
}

@Override
public void run() {
    latch.countDown();
    System.out.println("Count:- " + latch.getCount());
    System.out.println(Thread.currentThread().getName() + " finished");
}
}

输出为：-

数：-2

数：-1

数：-2

WORKER-3完成

WORKER-1完成

WORKER-2完成

计数：-0

最终计数：-0

WORKER-4完成了。

该计数在输出中两次返回2。我的代码有什么问题吗？

Answer 1

这不是问题。 request or session scope

此方法仅读取volatile，而不读取synchronised。

 /**
 * Returns the current count.
 *
 * <p>This method is typically used for debugging and testing purposes.
 *
 * @return the current count
 */
public long getCount() {
    return sync.getCount();
}

Answer 2

四个Worker同时运行。由于latch.countDown();和latch.getCount()不是原子的，因此执行顺序可能是：

thread1 latch.countDown();
thread2 latch.countDown();
thread1 System.out.println("Count:- " + latch.getCount()); // Count:- 2
thread2 System.out.println("Count:- " + latch.getCount()); // Count:- 2

Countdownlatch获取计数返回不一致的值

2 个答案: