Google Guava EventBus和事件处理程序中的异常

Question

Guava EventBus文档说 "一般情况下,处理程序不应抛出.如果这样做,EventBus将捕获并记录异常.这很少是错误处理的正确解决方案,不应该依赖;它仅用于帮助在开发过程中发现问题."

如果您知道可能发生某些异常,则可以使用EventBus注册SubscriberExceptionHandler并使用它来处理这些异常.

但是如果发生未处理的异常会发生什么？通常情况下,我希望一个未处理的异常"冒泡"调用链.使用SubscriberExceptionHandler时,我可以访问事件处理程序中抛出的原始异常,我只想重新抛出它.但我无法弄清楚如何.

那么,无论是否使用SubscriberExceptionHandler,如何确保事件处理程序中的意外异常不会被"吞噬"？

任何帮助将不胜感激.

Answer 1

番石榴不会让异常泡沫起来.它被迫在exceptionHandler中停止.请参阅下面的源代码.

      /**
       * Handles the given exception thrown by a subscriber with the given context.
       */
      void handleSubscriberException(Throwable e, SubscriberExceptionContext context) {
        checkNotNull(e);
        checkNotNull(context);
        try {
          exceptionHandler.handleException(e, context);
        } catch (Throwable e2) {
          // if the handler threw an exception... well, just log it
          logger.log(
              Level.SEVERE,
              String.format(Locale.ROOT, "Exception %s thrown while handling exception: %s", e2, e),
              e2);
        }
      }

我在github上发布了一个问题.您可以继承EventBus并编写自己的异常句柄逻辑.

package com.google.common.eventbus;

import com.google.common.eventbus.EventBus;
import com.google.common.eventbus.SubscriberExceptionContext;

/**
 * A eventbus wihch will throw exceptions during event handle process.
 * We think this behaviour is better.
 * @author ytm
 *
 */
public class BetterEventBus extends EventBus {

    public BetterEventBus() {}

    /**
     * Creates a new EventBus with the given {@code identifier}.
     *
     * @param identifier a brief name for this bus, for logging purposes. Should be a valid Java
     *     identifier.
     */
    public BetterEventBus(String identifier) {
        super(identifier);
    }

    /**
     * Just throw a EventHandleException if there's any exception.
     * @param e
     * @param context
     * @throws EventHandleException 
     */
    @Override
    protected void handleSubscriberException(Throwable e, SubscriberExceptionContext context) throws EventHandleException {
        throw new EventHandleException(e);
    }
}

Answer 2

如果要处理未检查的异常，则可以实现SubscriberExceptionHandler的方法，如下所示：

public void handleException(Throwable exception, SubscriberExceptionContext context) {
    // Check if the exception is of some type you wish to be rethrown, and rethrow it.
    // Here I'll assume you'd like to rethrow RuntimeExceptions instead of 'consuming' them.
    if (exception instanceof RuntimeException) {
        throw (RuntimeException) exception;
    }

    // If the exception is OK to be handled here, do some stuff with it, e.g. log it.
    ...
}

在创建实现SubscriberExceptionHandler接口的类之后，可以将其实例传递给EventBus的构造函数：

EventBus eventBus = new EventBus(new MySubscriberExceptionHandler());

完成后，eventBus将使用您的异常处理程序，该异常处理程序会RuntimeExceptions冒泡。