Java Puzzler：繁忙的等待线程停止工作

Question

这是某种Java拼图游戏，我偶然发现，无法真正解释。 也许有人可以吗？

短时间后，以下程序挂起。 有时在2个输出之后，有时在80个输出之后，但几乎总是在正确终止之前。 如果第一次没有发生，您可能必须运行几次。

public class Main {
    public static void main(String[] args) {
        final WorkerThread[] threads = new WorkerThread[]{ new WorkerThread("Ping!"), new WorkerThread("Pong!") };
        threads[0].start();
        threads[1].start();

        Runnable work = new Runnable() {
            private int counter = 0;
            public void run() {
                System.out.println(counter + " : " + Thread.currentThread().getName());
                threads[counter++ % 2].setWork(this);
                if (counter == 100) {
                    System.exit(1);
                }
            }
        };

        work.run();
    }
}

class WorkerThread extends Thread {
    private Runnable workToDo;

    public WorkerThread(String name) {
        super(name);
    }

    @Override
    public void run() {
        while (true){
            if (workToDo != null) {
                workToDo.run();
                workToDo = null;
            }
        }
    }

    public void setWork(Runnable newWork) {
        this.workToDo = newWork;
    }
}

现在，很明显，繁忙的等待循环通常不是一个好主意。 但这不是改善，而是了解正在发生的事情。

由于在synchronized WorkerThread.setWork()或将WorkerThread.workToDo字段设置为volatile时，一切都能按预期工作， WorkerThread.setWork()我怀疑存在内存问题。

但是为什么会这样呢？ 调试无济于事，一旦您开始逐步执​​行，一切都会按预期进行。

一个解释将不胜感激。

Answer 1

1. 第一个问题是您要从main线程设置Runnable workToDo ，然后在不同步的情况下在2个分支线程中读取它。 每当您在多个线程中修改一个字段时，都应将其标记为volatile或有人synchronized 。

    private volatile Runnable workToDo; 

2. 另外，由于多个线程正在执行counter++ ，所以这也需要synchronized 。 我为此推荐一个AtomicInteger 。

    private AtomicInteger counter = new AtomicInteger(0); ... threads[counter.incrementAndGet() % 2].setWork(this); 

3. 但是我认为真正的问题可能是比赛条件之一。 两个线程都可以将workToDo设置为Runnable ，然后让它们都返回并将其设置为null以便它们永远旋转。 我不确定该如何解决。

    1. threads[0] has it's `workToDo` set to the runnable. It calls `run()`. 2. at the same time threads[1] also calls `run()`. 3. threads[0] sets the `workToDo` on itself and threads[1] to be the runnable. 4. at the same time threads[1] does the same thing. 5. threads[0] returns from the `run()` method and sets `workToDo` to be `null`. 6. threads[1] returns from the `run()` method and sets `workToDo` to be `null`. 7. They spin forever... 

而且，正如您提到的那样，旋转循环很疯狂，但是我认为这是一个演示线程程序。

Answer 2

这些行之间直接发生问题：

workToDo.run();
workToDo = null;

假设发生以下事件序列：

- Original Runnable runs.  "Ping!".setWork() called
- Ping! thread realizes workToDo != null, calls run(), the stops between those two lines
  - "Pong!".setWork() called
- Pong! thread realizes workToDo != null, calls run()
  - "Ping!".setWork() called
- Ping! thread resumes, sets workToDo = null, ignorantly discarding the new value
- Both threads now have workToDo = null, and the counter is frozen at 2,...,80
  Program hangs

Answer 3

我的2美分...

import java.util.concurrent.atomic.AtomicReference;

class WorkerThread extends Thread {
    private AtomicReference<Runnable> work;

    public WorkerThread(String name) {
        super(name);
        work = new AtomicReference<Runnable>();
    }

    @Override
    public void run() {
        while (true){
            Runnable workToDo = work.getAndSet(null);
            if ( workToDo != null ) {
                workToDo.run();
            }
        }
    }

    public void setWork(Runnable newWork) {
        this.work.set(newWork);
    }
}