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Java ExecutorService invokeAll() 中断

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:28:41 27 4
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我有一个宽度为 10 的固定线程池 ExecutorService,以及 100 个 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对该列表调用 invokeAll,并且几乎立即中断了该线程。 ExecutorService执行如预期中断,但Callable记录的实际中断次数远超预期10次——20-40次左右。为什么会这样,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,您可能需要多次运行它)

@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

最佳答案

我不同意你应该只接收 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
....

从本质上讲,我的论点是不能保证 Watcher 线程在它必须产生时间片并允许 ExecutorService 的工作线程启动更多 Waiter 任务之前将被允许取消所有 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 block 是将中断传播到当前正在运行的任务的时间。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内遍历 Future 的所有实例。

关于Java ExecutorService invokeAll() 中断,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7976697/

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