RxJava flatMap和backpressure奇怪的行为

Question

在用RxJava编写数据同步作业时,我发现了一个我无法解释的奇怪行为.我是RxJava的新手,非常感谢帮助.

简而言之,我的工作非常简单我有一个元素ID列表,我调用web服务来获取每个元素的ID,进行一些处理并进行多次调用以将数据推送到DB.数据加载比数据存储更快,因此我输入了OutOfMemory错误.

我的代码几乎看起来像"失败"测试,但后来做了一些测试我意识到删除线:

flatMap(dt -> Observable.just(dt))

让它起作用.失败的测试输出清楚地表明未消耗的项目叠加,这导致OutOfMemory.工作测试输出显示生产者将始终等待消费者,因此这永远不会导致OutOfMemory.

public static class DataStore {
    public Integer myVal;
    public byte[] myBigData;

    public DataStore(Integer myVal) {
        this.myVal = myVal;
        this.myBigData = new byte[1000000];
    }
}

@Test
public void working() {
    int MAX_CONCURRENT_LOAD = 1;
    int MAX_CONCURRENT_STORE = 2;

    AtomicInteger nbUnconsumed = new AtomicInteger(0);

    List<Integer> ids = IntStream.range(0, 1000).boxed().collect(Collectors.toList());
    Observable.from(ids)
        .flatMap(this::produce, MAX_CONCURRENT_LOAD)
        .doOnNext(s -> logger.info("+1 Total unconsumed values: " + nbUnconsumed.incrementAndGet()))
        .flatMap(this::consume, MAX_CONCURRENT_STORE)
        .doOnNext(s -> logger.info("-1 Total unconsumed values: " + nbUnconsumed.decrementAndGet()))
        .toBlocking().forEach(s -> {});

    logger.info("Finished");
}

@Test
public void failing() {
    int MAX_CONCURRENT_LOAD = 1;
    int MAX_CONCURRENT_STORE = 2;

    AtomicInteger nbUnconsumed = new AtomicInteger(0);

    List<Integer> ids = IntStream.range(0, 1000).boxed().collect(Collectors.toList());
    Observable.from(ids)
        .flatMap(this::produce, MAX_CONCURRENT_LOAD)
        .doOnNext(s -> logger.info("+1 Total unconsumed values: " + nbUnconsumed.incrementAndGet()))
        .flatMap(dt -> Observable.just(dt))
        .flatMap(this::consume, MAX_CONCURRENT_STORE)
        .doOnNext(s -> logger.info("-1 Total unconsumed values: " + nbUnconsumed.decrementAndGet()))
        .toBlocking().forEach(s -> {});

    logger.info("Finished");
}

private Observable<DataStore> produce(final int value) {
    return Observable.<DataStore>create(s -> {
        try {
            if (!s.isUnsubscribed()) {
                Thread.sleep(200); //Here I synchronous call WS to retrieve data
                s.onNext(new DataStore(value));
                s.onCompleted();
            }
        } catch (Exception e) {
            s.onError(e);
        }
    }).subscribeOn(Schedulers.io());
}

private Observable<Boolean> consume(DataStore value) {
    return Observable.<Boolean>create(s -> {
        try {
            if (!s.isUnsubscribed()) {
                Thread.sleep(1000); //Here I synchronous call DB to store data
                s.onNext(true);
                s.onCompleted();
            }
        } catch (Exception e) {
            s.onNext(false);
            s.onCompleted();
        }
    }).subscribeOn(Schedulers.io());
}

这种行为背后的解释是什么？如何在不删除Observable.just(dt)的情况下解决我的失败测试,​​在我的实例中是Observable.from(someListOfItme)

Answer 1

flatMap 默认情况下,合并无限量的源并通过应用没有maxConcurrent参数的特定lambda,您基本上可以解除上行链接,现在可以全速运行,从而压倒其他运算符的内部缓冲区.