异步订阅服务器示例会丢失消息吗？

Question

public class SubscriberExample {

  private static final String PROJECT_ID = ServiceOptions.getDefaultProjectId();

  private static final BlockingQueue<PubsubMessage> messages = new LinkedBlockingDeque<>();

  static class MessageReceiverExample implements MessageReceiver {

    @Override
    public void receiveMessage(PubsubMessage message, AckReplyConsumer consumer) {
      messages.offer(message);
      consumer.ack();
    }
  }

  public static void main(String... args) throws Exception {
    String subscriptionId = args[0];
    ProjectSubscriptionName subscriptionName = ProjectSubscriptionName.of(
        PROJECT_ID, subscriptionId);
    Subscriber subscriber = null;
    try {
      subscriber =
          Subscriber.newBuilder(subscriptionName, new MessageReceiverExample()).build();
      subscriber.startAsync().awaitRunning();
      while (true) {
        PubsubMessage message = messages.take();
        processMessage(message);
      }
    } finally {
      if (subscriber != null) {
        subscriber.stopAsync();
      }
    }
  }

最好的实现难道不是只在消息被处理后才确认它吗？而不是确认它并将它留在队列中，并假设它将被处理。我明白这将解耦消息的接收和消息的处理，并潜在地增加吞吐量，但它仍然有丢失消息的风险，对吗？