Java ExecutorService invokeAll()中断

前端之家收集整理的这篇文章主要介绍了Java ExecutorService invokeAll()中断前端之家小编觉得挺不错的,现在分享给大家,也给大家做个参考。
我有一个宽度为10的固定线程池ExecutorService,以及一个100 Callable列表,每个等待20秒并记录它们的中断.

我在一个单独的线程中调用该列表中的invokeAll,几乎立即中断该线程. ExecutorService执行按预期中断,但Callables记录的实际中断数远远超过预期10 – 大约20-40.为什么这样,如果ExecutorService可以同时执行不超过10个线程?

完整来源:(由于并发,您可能需要运行一次)

@Test
public void interrupt3() throws Exception{
    int callableNum = 100;
    int executorThreadNum = 10;
    final AtomicInteger interruptCounter = new AtomicInteger(0);
    final ExecutorService executorService = Executors.newFixedThreadPool(executorThreadNum);
    final List <Callable <Object>> executeds = new ArrayList <Callable <Object>>();
    for (int i = 0; i < callableNum; ++i) {
        executeds.add(new Waiter(interruptCounter));
    }
    Thread watcher = new Thread(new Runnable() {

        @Override
        public void run(){
            try {
                executorService.invokeAll(executeds);
            } catch(InterruptedException ex) {
                // NOOP
            }
        }
    });
    watcher.start();
    Thread.sleep(200);
    watcher.interrupt();
    Thread.sleep(200);
    assertEquals(10,interruptCounter.get());
}

// This class just waits for 20 seconds,recording it's interrupts
private class Waiter implements Callable <Object> {
    private AtomicInteger    interruptCounter;

    public Waiter(AtomicInteger interruptCounter){
        this.interruptCounter = interruptCounter;
    }

    @Override
    public Object call() throws Exception{
        try {
            Thread.sleep(20000);
        } catch(InterruptedException ex) {
            interruptCounter.getAndIncrement();
        }
        return null;
    }
}

使用WinXP 32位,Oracle JRE 1.6.0_27和JUnit4

解决方法

我不同意你应该只收到10个中断的假设.
Assume the cpu has 1 core.
1. Main thread starts Watcher and sleeps
2. Watcher starts and adds 100 Waiters then blocks
3. Waiter 1-10 start and sleep in sequence
4. Main wakes and interrupts Watcher then sleeps
5. Watcher cancels Waiter 1-5 then is yielded by the OS   (now we have 5 interrupts)
6. Waiter 11-13 start and sleep
7. Watcher cancels Waiter 6-20 then is yielded by the OS   (now we have 13 interrupts)
8. Waiter 14-20 are "started" resulting in a no-op
9. Waiter 21-24 start and sleep
....

本质上,我的论点是,无法保证Watcher线程在允许生成时间片并允许ExecutorService的工作线程启动更多Waiter任务之前,将允许取消所有100个“Waiter”RunnableFuture实例.

更新:显示来自AbstractExecutorService的代码

public <T> List<Future<T>> invokeAll(Collection<? extends Callable<T>> tasks)
    throws InterruptedException {
    if (tasks == null)
        throw new NullPointerException();
    List<Future<T>> futures = new ArrayList<Future<T>>(tasks.size());
    boolean done = false;
    try {
        for (Callable<T> t : tasks) {
            RunnableFuture<T> f = newTaskFor(t);
            futures.add(f);
            execute(f);
        }
        for (Future<T> f : futures) {
            if (!f.isDone()) {
                try {
                    f.get(); //If interrupted,this is where the InterruptedException will be thrown from
                } catch (CancellationException ignore) {
                } catch (ExecutionException ignore) {
                }
            }
        }
        done = true;
        return futures;
    } finally {
        if (!done)
            for (Future<T> f : futures)
                f.cancel(true); //Specifying "true" is what allows an interrupt to be sent to the ExecutorService's worker threads
    }
}

包含f.cancel(true)的finally块是当中断传播到当前正在运行的任务时.如您所见,这是一个紧凑的循环,但不能保证执行循环的线程能够在一个时间片中遍历Future的所有实例.

猜你在找的Java相关文章