resolve #1645 | stop subscription immediately even when there are pen…

[pkasperowicz]

…ding retries

[faderskd]

@pkasperowicz are you sure we need this ? The stop() method is actually called in a finally block in the run() method. The main try block in the run() method is stopped when running.flag is set to false.

[pkasperowicz]

When the stop signal is received, we need to inform the consumer to stop consuming events.
stop() invokes consumer.tearDown() and it sets BatchConsumer.consuming = false.
It informs Retryer to stop retrying messages. Without that we need to wait for all reties to be finished and it is blocking that way the main loop.

Alternative approach might be extending the API of BatchConsumer to receive stop signal and mark it as some boolean e.g. stopSignalReceived. This flag would be informative for Retryer to stop the process of retries.
What do you think?

[faderskd]

ok I see it will work, can we avoid duplicated calling stop() method ? For example we can write method like this:

    private void stopWithExceptionHandling() {
        if (!running) {
            return;
        }
        try {
            stop();
        } catch (Exception exceptionWhileStopping) {
            logger.error("An error occurred while stopping consumer process of subscription {}",
                    getSubscriptionName(), exceptionWhileStopping);
        } finally {
            onConsumerStopped.accept(getSubscriptionName());
            Thread.currentThread().setName("consumer-released-thread");
        }
    }
``` and use it in both `finally` block of run method and in `process` method:
               

...
case STOP:
      logger.info("Stopping main loop for consumer {}. {}", signal.getTarget(), signal.getLogWithIdAndType());
      stopWithExceptionHandling();
      this.running = false;
      break;

[pkasperowicz]

@faderskd ok changed it

[faderskd]

@pkasperowicz thank you, +1