Question

我有以下两个类

public class TimeoutHandler
{
  private int timeoutMsec = 10000;
  private ScheduledExecutorService scheduler;
  private ScheduledFuture<String> future;

  public TimeoutHandler()
  {
  }

  public TimeoutHandler(int msec)
  {
    timeoutMsec = msec;
  }

  public void startTimeoutHandler() throws TimeoutException
  {
    scheduler = Executors.newScheduledThreadPool(1);
    future = scheduler.schedule(new TimeoutCallable(), timeoutMsec, TimeUnit.MILLISECONDS);
  }
}

和

public class TimeoutCallable implements Callable<String>
{
  @Override
  public String call() throws TimeoutException
  {
    throw new TimeoutException();
  }
}

现在我使用方法TimeoutHandler启动startTimeoutHandler()，它应该安排将来的线程从timeoutMsec开始。问题是，永远不会调用TimeoutCallable，并且永远不会启动throw new TimeoutException();。有什么建议吗？

Answer 1

我们需要调用future.get()来检索结果，如果以ExectuorException的形式提供，此函数将抛出异常：可以在异常上调用getCause()来检索原因。如下所示：

     try {
          future.get();
      } catch (InterruptedException ex) {
          Logger.getLogger(CallableTest.class.getName()).log(Level.SEVERE, null, ex);
      } catch (ExecutionException ex) {
          System.out.println(ex.getCause());
          //Logger.getLogger(CallableTest.class.getName()).log(Level.SEVERE, null, ex);
      }

<强>输出：

java.util.concurrent.TimeoutException

实现Callable的超时处理程序不会抛出异常

1 个答案: