LMAX Disruptor - 什么决定了批量大小？

Question

我最近一直在学习 LMAX Disruptor 并进行了一些实验。令我困惑的一件事是 EventHandler 的 onEvent 处理程序方法的 endOfBatch 参数。考虑我的以下代码。首先，我调用 Test1 和 Test1Worker:

的虚拟消息和消费者 classes

public class Test1 {

}

public class Test1Worker implements EventHandler<Test1>{
    public void onEvent(Test1 event, long sequence, boolean endOfBatch) {
        try{
            Thread.sleep(500);
        }
        catch(Exception e){
            e.printStackTrace();
        }
        System.out.println("Received message with sequence " + sequence + ". "
                + "EndOfBatch = " + endOfBatch);
    }
}

请注意，我已经设置了 500 毫秒的延迟，只是为了替代一些真实世界的工作。我还在控制台中打印

的序列号

然后我的 driver class（作为制作人）调用 DisruptorTest:

public class DisruptorTest {

    private static Disruptor<Test1> bus1;

    private static ExecutorService test1Workers;

    public static void main(String[] args){             
        test1Workers = Executors.newFixedThreadPool(1);

        bus1 = new Disruptor<Test1>(new Test1Factory(), 8, test1Workers);           
        bus1.handleEventsWith(new Test1Worker());
        RingBuffer<Test1> buf1 = bus1.start();

        for (int i = 0; i < 10; i++){
            long a = System.currentTimeMillis();
            long next = buf1.next();
            long b = System.currentTimeMillis();
            System.out.println("Delay for claiming slot " + i + " is "+ (b - a));
            try {
                Test1 message = buf1.get(next);
            } catch (Exception e) {
                e.printStackTrace();
            } finally {
                buf1.publish(next);
            }
        }
    }

    public static class Test1Factory implements EventFactory<Test1> {
        public Test1 newInstance() {
            return new Test1();
        }

    }   
}

在这里，初始化所需的东西后，我将 10 条消息提供给 RingBuffer（缓冲区大小 8）并尝试监视一些事情 - 生产者在中声明下一个插槽的延迟RingBuffer 和消息及其在消费者端的序列号，以及特定序列是否被视为批处理结束。

现在，有趣的是处理每条消息涉及 500 毫秒的延迟，这就是我得到的输出：

Delay for claiming slot 0 is 0
Delay for claiming slot 1 is 0
Delay for claiming slot 2 is 0
Delay for claiming slot 3 is 0
Delay for claiming slot 4 is 0
Delay for claiming slot 5 is 0
Delay for claiming slot 6 is 0
Delay for claiming slot 7 is 0
Received message with sequence 0. EndOfBatch = true
Delay for claiming slot 8 is 505
Received message with sequence 1. EndOfBatch = false
Received message with sequence 2. EndOfBatch = false
Received message with sequence 3. EndOfBatch = false
Received message with sequence 4. EndOfBatch = false
Received message with sequence 5. EndOfBatch = false
Received message with sequence 6. EndOfBatch = false
Received message with sequence 7. EndOfBatch = true
Delay for claiming slot 9 is 3519
Received message with sequence 8. EndOfBatch = true
Received message with sequence 9. EndOfBatch = true

但是，如果我删除 500 毫秒的等待时间，这就是我得到的：

Delay for claiming slot 0 is 0
Delay for claiming slot 1 is 0
Received message with sequence 0. EndOfBatch = true
Received message with sequence 1. EndOfBatch = true
Delay for claiming slot 2 is 0
Received message with sequence 2. EndOfBatch = true
Delay for claiming slot 3 is 0
Received message with sequence 3. EndOfBatch = true
Delay for claiming slot 4 is 0
Received message with sequence 4. EndOfBatch = true
Delay for claiming slot 5 is 0
Received message with sequence 5. EndOfBatch = true
Delay for claiming slot 6 is 0
Received message with sequence 6. EndOfBatch = true
Delay for claiming slot 7 is 0
Received message with sequence 7. EndOfBatch = true
Delay for claiming slot 8 is 1
Delay for claiming slot 9 is 0
Received message with sequence 8. EndOfBatch = false
Received message with sequence 9. EndOfBatch = true

所以看起来某个消息是否被认为是批处理的末尾（即，批处理的大小）受到消费者消息处理延迟的影响。可能是我在这里很愚蠢，但事情应该是这样吗？这背后的原因是什么？无论如何，通常是什么决定了批量大小？提前致谢。如果我的问题中有任何不清楚的地方，请告诉我。

Answer 1

批量大小仅由可用元素的数量决定。因此，如果此时有更多元素可用，那么它将包含在批次中。例如，如果 Disruptor 调用您的代码并且队列中只有一个元素，那么您将收到一个带有 endOfBatch=true 的调用。如果队列中有 8 个元素，那么它将收集所有 8 个元素并在一个批次中发送它们。

您可以在下面的代码中看到，队列中的条目数 "available" 已被提取，并且可能比 "next" 项多得多。所以比如你现在是5，在等slot 6，然后3个event到了，available会是8个，你会批量接到多个call（for 6,7,8）。

https://github.com/LMAX-Exchange/disruptor/blob/master/src/main/java/com/lmax/disruptor/BatchEventProcessor.java#L124

final long availableSequence = sequenceBarrier.waitFor(nextSequence);
while (nextSequence <= availableSequence)
{
    event = dataProvider.get(nextSequence);
    eventHandler.onEvent(event, nextSequence, nextSequence == availableSequence);
    nextSequence++;
}

关于元素 9 处的 500 毫秒暂停，请注意 Disruptor 是使用环形缓冲区构建的，并且您已将缓冲区中的槽数指定为 8（请参阅此处的第二个参数）：

bus1 = new Disruptor<Test1>(new Test1Factory(), 8, test1Workers);

如果不是所有消费者都消费了一个元素，并且环形缓冲区已满（所有 8 个元素都已满），生产者将被阻止向缓冲区发布新事件。您可以尝试增加缓冲区大小，比如 200 万个对象，或者确保您的消费者比生产者更快，这样队列就不会填满（删除睡眠，您已经证明了这一点）。

LMAX Disruptor - 什么决定了批量大小？

LMAX Disruptor - what determines the batch size?

java

multithreading

producer-consumer

disruptor-pattern

lmax