Question

我有一个宽度为10的固定线程池ExecutorService和一个100 Callable的列表，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，几乎立即中断了该线程。 ExecutorService执行按预期中断，但Callable s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？

完整源代码:(由于并发性，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

Answer 1

我不同意你应该只收到10次中断的假设。

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
....

基本上，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。 / p>

更新：显示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，但无法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。

Answer 2

如果你想达到同样的行为

    ArrayList<Runnable> runnables = new ArrayList<Runnable>();
    executorService.getQueue().drainTo(runnables);

在中断线程池之前添加此块。

它会将所有等待队列排入新列表。

因此它只会中断正在运行的线程。

Answer 3

PowerMock.mockStatic ( Executors.class );
EasyMock.expect ( Executors.newFixedThreadPool ( 9 ) ).andReturn ( executorService );

Future<MyObject> callableMock = (Future<MyObject>) 
EasyMock.createMock ( Future.class );
EasyMock.expect ( callableMock.get ( EasyMock.anyLong (), EasyMock.isA ( TimeUnit.class ) ) ).andReturn ( ccs ).anyTimes ();

List<Future<MyObject>> futures = new ArrayList<Future<MyObject>> ();
futures.add ( callableMock );
EasyMock.expect ( executorService.invokeAll ( EasyMock.isA ( List.class ) ) ).andReturn ( futures ).anyTimes ();

executorService.shutdown ();
EasyMock.expectLastCall ().anyTimes ();

EasyMock.expect ( mock.getMethodCall ( ) ).andReturn ( result ).anyTimes ();

PowerMock.replayAll ();
EasyMock.replay ( callableMock, executorService, mock );

Assert.assertEquals ( " ", answer.get ( 0 ) );
PowerMock.verifyAll ();

Java ExecutorService invokeAll（）中断

3 个答案: