4

Why does this code not print an exception stack trace?

public class Playground {

    /**
     * @param args
     */
    public static void main(String[] args) {
        startThread();
    }

    private static void startThread() {
        ScheduledExecutorService timer = Executors
                .newSingleThreadScheduledExecutor();
        Runnable r = new Runnable() {
            int dummyInt = 0;
            boolean dummyBoolean = false;

            @Override
            public void run() {
                dummyInt = Integer.parseInt("AAAA");

                if (dummyBoolean) {
                    dummyBoolean= false;
                } else {
                    dummyBoolean= true;
                }

            }

        };

        timer.scheduleAtFixedRate(r, 0, 100, TimeUnit.MILLISECONDS);

    }

}

How can I get it to?

I would expect to see this:

java.lang.NumberFormatException: For input string: "AAAA"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at Playground$1.run(Playground.java:25)
    at java.util.concurrent.Executors$RunnableAdapter.call(Unknown Source)
    at java.util.concurrent.FutureTask$Sync.innerRunAndReset(Unknown Source)
    at java.util.concurrent.FutureTask.runAndReset(Unknown Source)
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$101(Unknown Source)
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.runPeriodic(Unknown Source)
    at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.runTask(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at java.lang.Thread.run(Unknown Source)
4

3 回答 3

8

执行程序可能会在线程上设置自己的未捕获异常处理程序，因此堆栈跟踪不会打印到控制台。如果在中抛出异常，Runnable您可以从方法ScheduledFuture返回的对象中获取它scheduleAtFixedRate：

ScheduledFuture<?> future = timer.scheduleAtFixedRate(r, 0, 100, TimeUnit.MILLISECONDS);
try {
    future.get();
} catch (ExecutionException e) {
    Throwable cause = e.getCause();
    cause.printStackTrace();
}
于 2010-03-17T00:55:20.010 回答
1

由于某种原因，执行器服务倾向于吞下异常。它可以使调试变得非常具有挑战性。只需在 run 方法的内容周围添加一个 try catch，如下所示：

public void run() {
  try
    dummyInt = Integer.parseInt("AAAA");

    if (dummyBoolean) {
      dummyBoolean= false;
    } else {
      dummyBoolean= true;
    }
  } catch (Exception e){
    e.printStackTrace();
  }

}

这是一个相关的问题。

于 2010-03-17T00:56:56.247 回答
0

在 StackOverflow 上有一个类似的问题提供了一个答案Unhandled exceptions with Java scheduled executors here。

基本上你可以在scheduleAtFixedRate() javadoc中阅读：

如果任务的任何执行遇到异常，后续执行将被抑制

所以，我从这里概括一下，在通过ScheduledExecutorService.scheduleAtFixedRate().

于 2010-03-17T00:56:21.897 回答