Java中导致超时异常时如何获取Future返回 [英] how to get Future return when cause timeout Exception in java

问题描述

我想知道当使用Future和ExecutorService时，如果导致超时异常会导致我得到结果，例如，下面是我的代码

I'm wondering can I get the result when it cause timeout Exception when use Future and ExecutorService, for example, below is my code

package com.example;

import java.util.UUID;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Future;
import java.util.concurrent.LinkedBlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.TimeoutException;

import org.apache.commons.lang3.concurrent.BasicThreadFactory;
import org.junit.Test;

public class FutureTest {
    public static String sayHello() {
        try {
            Thread.sleep(3000);
            return "world";
        } catch (Exception e) {
            return "hello";
        }
    }

    @Test
    public void testTimeoutAndInterrupt() throws Exception {
        ExecutorService executorService = new ThreadPoolExecutor(0, Runtime.getRuntime().availableProcessors() * 2, 0,
            TimeUnit.SECONDS, new LinkedBlockingQueue<>(10), new BasicThreadFactory.Builder().namingPattern(
            "executor-pool-%d").daemon(true).build());

        String processName = UUID.randomUUID().toString();

        Future<String> future = executorService.submit(
            () -> {
                return sayHello();
            }
        );
        try {
            String result = future.get(2, TimeUnit.SECONDS);
            System.out.println("future result:" + result);
        } catch (TimeoutException e) {
            System.out.println("timeout exception");
        } finally {
            future.cancel(true);
            System.out.println("cancel by future");
        }
    }
}

当我使用future.get(5，TimeUnit.SECONDS)时，可以按预期获得值世界".现在，当我使用future.get(2，TimeUnit.SECONDS)时，我想要返回的值是"hello".

when I use future.get(5, TimeUnit.SECONDS), It's OK to get value "world" as I expect. now I want get the return value is "hello", when I use future.get(2, TimeUnit.SECONDS) .

我该怎么做?谢谢

推荐答案

不，该异常不会在执行过程中抛出.而是抛出一个 InterruptedException .(在那里尝试使用System.out.println).

No, that exception is not thrown inside the execution. Instead a InterruptedException is thrown. (Try a System.out.println there).

控制流程:

1. 执行中发生超时

1. 线程被中断

3a.抛出TimeoutException

3b.FutureTask收到InterruptedException

当3a和3b在两个线程中(异步)发生时，顺序可能是3b; 3a.但是3a; 3b似乎是自然顺序.

As 3a and 3b happen in two threads (asynchroneously) the order could be 3b;3a. However 3a;3b seems the natural order.

这样做:

    String result;
    try {
        result = future.get(2, TimeUnit.SECONDS);
        System.out.println("future result:" + result);
    } catch (TimeoutException e) {
        System.out.println("timeout exception");
        result = "hello";
    } finally {

注意:我认为FutureTask线程不能保证已关闭(因为可能没有捕获，即使捕获之后代码仍会继续).

Notice: I think the FutureTask thread is not guaranteed to have shutdown (as there might not have been a catch, and even then after the catch the code continues).

将catch InterruptedException捕获到某个地方存放结果将失败，因为在捕获TimeoutException时，FutureTask仍然不能具有InterruptedException.

Using catch InterruptedException to somewhere deposit a result, will fail, as at the moment of catching the TimeoutException the FutureTask still can have no InterruptedException.

在每个循环步骤中存储进度时，这也是一个陷阱.捕获超时后，进度可能仍会更新.

This is also a pitfall when storing a progress in every loop step. After catching the timeout the progress might still be updated.

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