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异常没有输出

executorservice exception multithreading java

mainstringargs · 技术社区 · 14 年前

为什么此代码不打印异常堆栈跟踪?

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);

    }

}

我怎样才能拿到?

我希望看到:

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)

3 回复 | 直到 14 年前

Kevin K 14 年前

执行器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 Runnable 你可以从 ScheduledFuture 返回的对象 scheduleAtFixedRate 方法:

ScheduledFuture<?> future = timer.scheduleAtFixedRate(r, 0, 100, TimeUnit.MILLISECONDS);
try {
    future.get();
} catch (ExecutionException e) {
    Throwable cause = e.getCause();
    cause.printStackTrace();
}

Community leo1 7 年前

由于某种原因,执行器服务倾向于接受异常。它可以使调试非常具有挑战性。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();
  }

}

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