ExecutorService 使用 invokeAll 和超时异常后可调用线程上的超时未终止

Posted 2023-02-25

【中文标题】ExecutorService 使用 invokeAll 和超时异常后可调用线程上的超时未终止

【英文标题】：ExecutorService using invokeAll and timeout on callables thread not got terminated after timeout exception

【发布时间】：2021-08-27 08:22:58

【问题描述】：

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，但这不是超时工作发现的情况 但线程完成执行。 如果它通过超时，我如何终止线程？ 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();
        
    

【参考方案1】：

您的Thread 在超过超时后没有完成它的执行。

请看这个例子作为参考：

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();

【讨论】：

谢谢，我也知道 executorService 永远不会抛出 TimeoutException ，我怎么知道它是超时的？来自线程的中断异常。睡眠不是来自 executorService 它自己

@user63898 invokeAll 不会抛出 TimeoutException。如果您想知道未来是否在完成执行之前超时，那么您必须通过它的第二个 get-Method 直接询问 Future。 -> V get(long timeout, TimeUnit unit)

我喜欢从 TimeoutException 类型中捕获异常，以便我有指标。我注意到这是我做的： executor.submit(new Callable() public String call() throws Exception exec(ac); return "done"; ).get(this.threadTimeOutInSec, TimeUnit.SECONDS );它确实从 TimeoutException 抛出了正确的异常 invokeAll 和一一提交有什么区别？

@user63898 简单地说：这是两种不同的实现方式。 invokeAll 允许您一次提交多个Callables。如果您现在通过invokeAll 执行它们或逐个提交它们，则取决于您作为开发人员。

谢谢我喜欢调用不阻塞的东西