在超时异常后未终止可调用线程上使用 invokeAll 和超时的 ExecutorService

Question

我尝试在线程上设置超时并期望执行程序抛出异常并阻止线程形式运行它终止它但超时工作发现情况并非如此但线程完成执行。如果超过超时，我该如何终止线程？这是我的测试代码：

class ArithmeticBB implements ArithmeticManagerCallable.ArithmeticAction {
    @Override
    public String arithmetic(String n) {
        try {
            Thread.sleep(5000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        String ss  = n+" 2" + " ,Thread ID:" +Thread.currentThread().getId();
        return ss;
    }
}
public class ArithmeticManagerCallable {
    ExecutorService executor = null;
    private List<String> integerList = null;
    private List<String> myResult= Collections.synchronizedList(new ArrayList<>());
    private int threadTimeOutInSec = 180;

    public ArithmeticManagerCallable(List<String> dataFromUser, int poolSize, int threadTimeOutInSec) {
        this.integerList =  dataFromUser;
        executor = Executors.newFixedThreadPool(poolSize);
        this.threadTimeOutInSec = threadTimeOutInSec;
    }
    private void exec(ArithmeticAction arithmeticAction) {
        List<String> tempList = new ArrayList<>();
        for(Iterator<String> iterator = integerList.listIterator(); iterator.hasNext();) {
            tempList.add(arithmeticAction.arithmetic(iterator.next()));
        }
        resultArray.addAll(tempList);
    }
    public List<String> invokerActions(List<ArithmeticAction> actions) throws
            InterruptedException {

        Set<Callable<String>> callables = new HashSet<>();
        for (final ArithmeticAction ac : actions) {
            callables.add(new Callable<String>() {
                public String call() throws Exception{
                    exec(ac);
                    return "done";
                }
            });
        }
        List<Future<String>> futures = executor.invokeAll(callables, this.threadTimeOutInSec, TimeUnit.SECONDS);
        executor.shutdown();
        while (!executor.isTerminated()) {
        }
        return myResult;
    }
    public interface ArithmeticAction {
        String arithmetic(String n);
    }

    public static void main(String[] args) {
        List<ArithmeticManagerCallable.ArithmeticAction> actions = new ArrayList();
        actions.add(new ArithmeticBB());
        List<String> intData = new ArrayList<>();
        intData.add("1");

        ArithmeticManagerCallable arithmeticManagerCallable = new ArithmeticManagerCallable(intData,20,4);

        try {
            List<String> result = arithmeticManagerCallable.invokerActions(actions);
            System.out.println("***********************************************");
            for(String i : result) {
                System.out.println(i);
            }
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

Answer 1

您的 Thread 在超过超时后未完成执行。

参考这个例子：

ExecutorService executorService = Executors.newFixedThreadPool(20);

List<Callable<String>> callableList = new ArrayList<>();
for (int i = 0; i < 10; i++) {
    final int identifier = i;
    callableList.add(()-> {
        try {
            Thread.sleep(1000 * identifier);
        } catch (InterruptedException e) {
            System.out.println("I'm " + identifier + " and my sleep was interrupted.");
        }
        return identifier + " Hello World";
    });
}

try {
    List<Future<String>> futureList = executorService.invokeAll(callableList, 5, TimeUnit.SECONDS);

    for (Future<String> result : futureList) {
        System.out.println(result.get());
    }
} catch (InterruptedException | ExecutionException e) {
    System.out.println("Something went wrong while executing. This may help: " + e.getMessage());
} catch (CancellationException e) {
    System.out.println("Not all Futures could be received.");
} finally {
    executorService.shutdown();
}

在超时异常后未终止可调用线程上使用 invokeAll 和超时的 ExecutorService

ExecutorService using invokeAll and timeout on callables thread not got terminated after timeout exception

java

multithreading

timeout

executorservice

threadpool