Java——线程执行顺序

Question

我正在尝试使用信号量严格依次启动 10 个线程。 也就是说，在thread-0执行之后，应该执行thread-1，而不是thread-2。

但问题是线程无序到达semaphore.acquire()-方法，因此线程的执行是无序的。 如何使用信号量而不使用thread.join() 来解决这个问题？

public class Main {

    private Semaphore semaphore = new Semaphore(1, true);

    public static void main(String[] args) {
        new Main().start();
    }

    private void start() {
        for (int i = 0; i < 10; i++) {
            Runnable runnable = () -> {
                try {
                    semaphore.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }

                System.out.println("In run method " + Thread.currentThread().getName());
                semaphore.release();
            };
            Thread thread = new Thread(runnable);
            thread.start();
        }
    }
    
}

输出：

In run method Thread-0
In run method Thread-1
In run method Thread-4
In run method Thread-5
In run method Thread-3
In run method Thread-2
In run method Thread-6
In run method Thread-7
In run method Thread-9
In run method Thread-8

Answer 1

您需要一个具有某种排序概念的同步对象。如果您熟悉美国杂货店，请考虑熟食柜台上的“取号”设备，它会告诉您轮到谁了。

代码粗略：

class SyncThing {
   int turn = 0; 
   synchronized void waitForTurn(int me) {
       while (turn != me)
           wait();
   }
   synchronized void nextTurn() {
        turn++;
        notifyAll();
   }
}

然后声明SyncThing syncThing = new SyncThing();

然后运行第 i 个线程：

        Runnable runnable = () -> {
            syncThing.waitForTurn(i);
            System.out.println("In run method " + Thread.currentThread().getName());
            syncThing.nextTurn();
        };

这是在我的脑海中输入的，并未作为完整的代码提供，但它应该显示方式。

Answer 2

private void start() {
    final AtomicInteger counter = new AtomicInteger();

    for (int i = 0; i < 10; i++) {
        final int num = i;
        new Thread(() -> {
            while (counter.get() != num) {
            }
            System.out.println("In run method " + Thread.currentThread().getName());
            counter.incrementAndGet();
        }).start();
    }
}