如何使用invokeAll()让所有线程池完成他们的任务?

时间:2013-08-13 06:46:50

标签: java multithreading arraylist executorservice future

    ExecutorService pool=Executors.newFixedThreadPool(7);
        List<Future<Hotel>> future=new ArrayList<Future<Hotel>>();
        List<Callable<Hotel>> callList = new ArrayList<Callable<Hotel>>();

        for(int i=0;i<=diff;i++){

            String str="2013-"+(liDates.get(i).get(Calendar.MONTH)+1)+"-"+liDates.get(i).get(Calendar.DATE);

            callList.add(new HotelCheapestFare(str));

        }       
     future=pool.invokeAll(callList);
for(int i=0;i<=future.size();i++){

        System.out.println("name is:"+future.get(i).get().getName());
    }

现在我希望在进入for循环之前将池invokeAll放到invokeAll所有任务,但是当我运行此循环的程序时,在java.util.concurrent.ExecutionException: java.lang.NullPointerException at java.util.concurrent.FutureTask$Sync.innerGet(Unknown Source) at java.util.concurrent.FutureTask.get(Unknown Source) at com.mmt.freedom.cheapestfare.TestHotel.main(TestHotel.java:6‌​5) Caused by: java.lang.NullPointerException at com.mmt.freedom.cheapestfare.HotelCheapestFare.getHotelCheap‌estFare(HotelCheapes‌​tFare.java:166) at com.mmt.freedom.cheapestfare.HotelCheapestFare.call(HotelChe‌​apestFare.java:219) at com.mmt.freedom.cheapestfare.HotelCheapestFare.call(HotelChe‌​apestFare.java:1) at java.util.concurrent.FutureTask$Sync.innerRun(Unknown Source) at java.util.concurrent.FutureTask.run(Unknown Source) at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source) atjava.util.concurrent.ThreadPoolExecutor$Worker.run(Unknow‌​n Source) at java.lang.Thread.run 之前执行并抛出此异常:

{{1}}

3 个答案:

答案 0 :(得分:20)

ExecutorService的工作方式是,当您致电invokeAll时,它会等待所有任务完成:

  

执行给定的任务,返回持有他们的期货清单   完成后的状态和结果。 Future.isDone()对每个都是正确的   返回列表的元素。 请注意,已完成的任务可能有   正常终止或通过抛出异常终止。结果   如果同时修改给定集合,则此方法未定义   此操作正在进行中。 1 (强调添加)

这意味着你的任务都已完成,但有些人可能会抛出异常。此异常是Future的一部分 - 调用get会导致异常重新包装在ExecutionException中。

来自你的stacktrack

java.util.concurrent.ExecutionException: java.lang.NullPointerException at
java.util.concurrent.FutureTask$Sync.innerGet(Unknown Source) at
java.util.concurrent.FutureTask.get(Unknown Source) at 
                                ^^^ <-- from get

你可以看到确实如此。您的一项任务因NPE而失败。 ExecutorService抓住了异常,并在您致电ExecutionException时抛出Future.get来告诉您。

现在,如果您想完成任务,则需要ExecutorCompletionService。这充当BlockingQueue,允许您在完成任务时轮询任务。

public static void main(String[] args) throws Exception {
    final ExecutorService executorService = Executors.newFixedThreadPool(10);
    final ExecutorCompletionService<String> completionService = new ExecutorCompletionService<>(executorService);
    executorService.submit(new Runnable() {
        @Override
        public void run() {
            for (int i = 0; i < 100; ++i) {
                try {
                    final Future<String> myValue = completionService.take();
                    //do stuff with the Future
                    final String result = myValue.get();
                    System.out.println(result);
                } catch (InterruptedException ex) {
                    return;
                } catch (ExecutionException ex) {
                    System.err.println("TASK FAILED");
                }
            }
        }
    });
    for (int i = 0; i < 100; ++i) {
        completionService.submit(new Callable<String>() {
            @Override
            public String call() throws Exception {
                if (Math.random() > 0.5) {
                    throw new RuntimeException("FAILED");
                }
                return "SUCCESS";
            }
        });
    }
    executorService.shutdown();
}

在此示例中,我有一个任务在take上调用ExecutorCompletionService,当Future变为可用时获取ExecutorCompletionService,然后我将任务提交给Future

这将允许您在失败时立即获取失败的任务,而不必等待所有任务一起失败。

唯一的复杂因素是很难告诉轮询线程所有任务都已完成,因为现在所有任务都是异步的。在这个例子中,我使用了提交100个任务的知识,因此它只需要轮询100次。更通用的方法是从submit方法收集{{1}},然后循环遍历它们以查看是否所有内容都已完成。

答案 1 :(得分:4)

Future.get()会抛出异常。

CancellationException - 如果计算被取消

ExecutionException - 如果计算引发异常

InterruptedException - 如果当前线程在等待时被中断

调用get()方法时,请抓住所有这些例外情况。

我已为某些Callable任务模拟除零例外,但如果您遇到上述三个例外情况,则Callable中的例外不会影响提交给Callable的其他ExecutorService任务如示例代码所示。

示例代码段:

import java.util.concurrent.*;
import java.util.*;

public class InvokeAllUsage{
    public InvokeAllUsage(){
        System.out.println("creating service");
        ExecutorService service = Executors.newFixedThreadPool(10);

        List<MyCallable> futureList = new ArrayList<MyCallable>();
        for ( int i=0; i<10; i++){
            MyCallable myCallable = new MyCallable((long)i+1);
            futureList.add(myCallable);
        }
        System.out.println("Start");
        try{
            List<Future<Long>> futures = service.invokeAll(futureList);  
            for(Future<Long> future : futures){
                try{
                    System.out.println("future.isDone = " + future.isDone());
                    System.out.println("future: call ="+future.get());
                }
                catch (CancellationException ce) {
                    ce.printStackTrace();
                } catch (ExecutionException ee) {
                    ee.printStackTrace();
                } catch (InterruptedException ie) {
                    Thread.currentThread().interrupt(); // ignore/reset
                }
            }
        }catch(Exception err){
            err.printStackTrace();
        }
        System.out.println("Completed");
        service.shutdown();
    }
    public static void main(String args[]){
        InvokeAllUsage demo = new InvokeAllUsage();
    }
    class MyCallable implements Callable<Long>{
        Long id = 0L;
        public MyCallable(Long val){
            this.id = val;
        }
        public Long call(){

            if ( id % 5 == 0){
                id = id / 0;
            }           
            return id;
        }
    }
}

输出:

creating service
Start
future.isDone = true
future: call =1
future.isDone = true
future: call =2
future.isDone = true
future: call =3
future.isDone = true
future: call =4
future.isDone = true
java.util.concurrent.ExecutionException: java.lang.ArithmeticException: / by zero
        at java.util.concurrent.FutureTask.report(FutureTask.java:122)
        at java.util.concurrent.FutureTask.get(FutureTask.java:188)
        at InvokeAllUsage.<init>(InvokeAllUsage.java:20)
        at InvokeAllUsage.main(InvokeAllUsage.java:37)
Caused by: java.lang.ArithmeticException: / by zero
        at InvokeAllUsage$MyCallable.call(InvokeAllUsage.java:47)
        at InvokeAllUsage$MyCallable.call(InvokeAllUsage.java:39)
        at java.util.concurrent.FutureTask.run(FutureTask.java:262)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
        at java.lang.Thread.run(Thread.java:744)
future.isDone = true
future: call =6
future.isDone = true
future: call =7
future.isDone = true
future: call =8
future.isDone = true
future: call =9
future.isDone = true
java.util.concurrent.ExecutionException: java.lang.ArithmeticException: / by zero
        at java.util.concurrent.FutureTask.report(FutureTask.java:122)
        at java.util.concurrent.FutureTask.get(FutureTask.java:188)
        at InvokeAllUsage.<init>(InvokeAllUsage.java:20)
        at InvokeAllUsage.main(InvokeAllUsage.java:37)
Caused by: java.lang.ArithmeticException: / by zero
        at InvokeAllUsage$MyCallable.call(InvokeAllUsage.java:47)
        at InvokeAllUsage$MyCallable.call(InvokeAllUsage.java:39)
        at java.util.concurrent.FutureTask.run(FutureTask.java:262)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
        at java.lang.Thread.run(Thread.java:744)
Completed

答案 2 :(得分:-1)

invokeAll是一种阻止方法。这意味着 - 在完成所有线程之前,JVM不会进入下一行。所以我认为你的线程未来结果有问题。

System.out.println("name is:"+future.get(i).get().getName());

从这一行我认为有一些期货没有结果可以为null,所以你应该检查你的代码,如果有一些Futures null,如果是这样的话,在执行这行之前得到一个if。