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2.x: Observable.combineLatest skips async events

See original GitHub issue

Versions

RxJava: 2.2.4
RxAndroid: 2.1.0

Sample

val numberStream = mainThreadObservable // emits something
    .map { 42 }
    .logOnNext("source: number")
val booleanStream = ioThreadObservable // emits "true" at undefined moment
    .startWith(false)
    .logOnNext("source: boolean")

disposable += Observable
    .combineLatest(
        numberStream.logOnNext("combination: number"),
        booleanStream.logOnNext("combination: boolean")
    )
    .logOnNext("combination")
    .subscribe()

Output

Actual

13:06:44.654 I: [main] [source: boolean]: false
13:06:44.658 I: [main] [combination: number]: 42
13:06:44.660 I: [main] [combination: boolean]: false
13:06:44.661 I: [Rx-io-20] [source: boolean]: true
13:06:44.667 I: [main] [combination]: (42, false)

Expected

Everything as in Actual, but with this line in the end.

13:06:44.667 I: [main] [combination]: (42, true)

Thoughts

This does not happen all the time consistently though. Researching this behavior lead me to #5325. Is it true that Observable.combineLatest is not guaranteed to process all onNext events when dealing with multi-thread streams? What is the best practice in this scenario? Add observeOn(Schedulers.single()) to all streams in combineLatest statement?

Issue Analytics

State:
Created 5 years ago
Reactions:2
Comments:8 (8 by maintainers)

Top GitHub Comments

1reaction

arturdryomovcommented, Dec 14, 2018

Damn, the referred issue was created 4 hours ago. I swear it wasn’t on purpose 😉

I tend to disagree though. If I’m reading the article correctly André describes a case when both combinations are emitted. The sample I’ve attached above skips one entirely. I agree that the ordering might be lost due to async nature, but skipping the event looks like another issue.

0reactions

arturdryomovcommented, Dec 14, 2018

Yep, it is never here. I’ll do my best ASAP, can be tricky due to multithreading though, but let’s see how it goes. Thanks!

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