Java-线程执行顺序

问题描述

我正尝试使用信号灯严格依次启动10个线程。 也就是说,在执行线程0之后,应该执行线程1,而不是线程2。

但是问题在于,线程是按乱序到达semaphore.acquire()方法的,因此线程的执行是乱序的。 如何使用信号量但不使用thread.join()解决此问题?

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

解决方法

您需要具有某种排序概念的同步对象。如果您熟悉美国的杂货店,请考虑在熟食店的柜台上“取个号码”,告诉您轮到谁了。

粗略的代码:

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();

并因此运行第i个线程:

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

这是我脑海中输入的,没有提供完整的代码,但应该可以显示出方法。

,
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();
    }
}