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Java Executors: how can I stop submitted tasks?

I have submitted a task using executors and I need it to stop after some time (e.g. 5 minutes). I have tried doing like this:

   for (Future<?> fut : e.invokeAll(tasks, 300, TimeUnit.SECONDS)) {
         try {
             fut.get(); 
         } catch (CancellationException ex) {
             fut.cancel(true);   
             tasks.clear();
         } catch(ExecutionException ex){
             ex.printStackTrace(); //FIXME: gestita con printstack       
         }
   }

But I always get an error: I have a shared Vector that needs to be modified by the tasks and then read by a thread, and even if I stop all the task, if the timeout occurs I get:

Exception in thread "Thread-1" java.util.ConcurrentModificationException

Is there something wrong? How can I stop the tasks submitted that are still working after 5 minutes?

like image 644
Raffo Avatar asked Sep 13 '09 15:09

Raffo


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1 Answers

Just because you call cancel() on Future doesn't mean that the task will stop automatically. You have to do some work within the task to make sure that it will stop:

  • Use cancel(true) so that an interrupt is sent to the task.
  • Handle InterruptedException. If a function in your task throws an InterruptedException, make sure you exit gracefully as soon as possible upon catching the exception.
  • Periodically check Thread.currentThread().isInterrupted() if the task does continuous computation.

For example:

class LongTask implements Callable<Double> {
    public Double call() {
        
         // Sleep for a while; handle InterruptedException appropriately
         try {
             Thread.sleep(10000);
         } catch (InterruptedException ex) {
             System.out.println("Exiting gracefully!");
             return null;
         }


         // Compute for a while; check Thread.isInterrupted() periodically
         double sum = 0.0;
         for (long i = 0; i < 10000000; i++) {
             sum += 10.0
             if (Thread.currentThread().isInterrupted()) {
                 System.out.println("Exiting gracefully");
                 return null;
             }
         }

         return sum;
    } 
}

Also, as other posts have mentioned: ConcurrentModificationException can be thrown even if using the thread-safe Vector class, because iterators you obtain from Vector are not thread-safe, and thus need to be synchronized. The enhanced for-loop uses iterators, so watch out:

final Vector<Double> vector = new Vector<Double>();
vector.add(1.0);
vector.add(2.0);

// Not thread safe!  If another thread modifies "vector" during the loop, then
// a ConcurrentModificationException will be thrown.
for (Double num : vector) {
    System.out.println(num);
}

// You can try this as a quick fix, but it might not be what you want:
synchronized (vector) {    // "vector" must be final
    for (Double num : vector) {
        System.out.println(num);
    }
}
like image 149
Matt Fichman Avatar answered Sep 27 '22 18:09

Matt Fichman