Why doesn't subscribeOn effect on PublishSubject in Rxjava?

Question

This is my test code in Kotlin:

fun main() {
    rxjava()
}

fun rxjava() {
    val queuSubject = PublishSubject.create<String>()
    queuSubject
        .map { t ->
            val a = t.toLong()
            Thread.sleep(6000 / a)
            println("map $a called ${Thread.currentThread().name} ")
            a
        }
        .subscribeOn(Schedulers.io())
        .observeOn(Schedulers.io())
        .subscribe({
            println("thread in subscription ${Thread.currentThread().name}")
        }, {
            println("error ${it.message}")
        })
    for (i in 1..3) {
        Thread {
            queuSubject.onNext("$i")
        }.start()
    }
    Thread.sleep(15000)
}

I'm trying to run map block and subscribe's onNext block in different IO threads. But the output is like this:

map 3 called Thread-2 
thread in subscription RxCachedThreadScheduler-2
map 2 called Thread-1 
thread in subscription RxCachedThreadScheduler-2
map 1 called Thread-0 
thread in subscription RxCachedThreadScheduler-2

As you can see It seems that calling subscribeOn has no effect on PublishSubject's stream and thread-0,thread-1 and thread-2 refers to the threads that call onNext methods.

Additionally consider the code below:

fun main() {
    rxjava()
}

fun rxjava() {
    val queuSubject = PublishSubject.create<String>()
    queuSubject
        .map { t ->
            val a = t.toLong()
            Thread.sleep(6000 / a)
            println("map $a called ${Thread.currentThread().name} ")
            a
        }
        .subscribeOn(Schedulers.io())
        .observeOn(Schedulers.io())
        .subscribe({
            println("thread in subscription ${Thread.currentThread().name}")
        }, {
            println("error ${it.message}")
        })
    queuSubject.onNext("1")
    queuSubject.onNext("2")
    queuSubject.onNext("3")
    Thread.sleep(15000)
}

I wrote the code above and saw that no output is printed. But If I remove subscribeOn from the stream, messages are printed sequentially like the following:

map 1 called main 
thread in subscription RxCachedThreadScheduler-1
map 2 called main 
thread in subscription RxCachedThreadScheduler-1
map 3 called main 
thread in subscription RxCachedThreadScheduler-1

What is the problem of these codes? Thanks.

score 3 · Accepted Answer · answered Jul 02 '21 at 08:42

3

Because subscribeOn only affects subscription side-effects of a source. Such side-effect would be if the source starts emitting events right when an observer subscribes:

Observable.just(1, 2, 3)
.subscribeOn(Schedulers.io())
.doOnNext(v -> System.out.println(Thread.currentThread() + " - " + v)
.blockingSubscribe();

PublishSubject has no subscription side-effect as it only relays signals from its onXXX method to the observers' onXXX methods.

However, subscribeOn has a time effect as it delays the actual subscription to the source, thus in case of PublishSubject, it might not see a registered observer in time some other thread calls its onXXX methods.

If you want to move the processing off of the original thread, use observeOn:

val queuSubject = PublishSubject.create<String>()
    queuSubject
        .observeOn(Schedulers.io()) // <----------------------------------------
        .map { t ->
            val a = t.toLong()
            Thread.sleep(6000 / a)
            println("map $a called ${Thread.currentThread().name} ")
            a
        }
        .observeOn(Schedulers.io())
        .subscribe({
            println("thread in subscription ${Thread.currentThread().name}")
        }, {
            println("error ${it.message}")
        })

answered Jul 02 '21 at 08:42

akarnokd

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I don't know why but the code you wrote prints only the message of one of onNext calls and not the rest. I mean just `map 1 called RxCachedThreadScheduler-1 ` and `thread in subscription RxCachedThreadScheduler-2` is printed. – Hadi Jul 02 '21 at 09:02
Did you include the Thread.sleep? How did you feed it values? Your first example is invalid as you should not call `onNext` from multiple threads concurrently. – akarnokd Jul 02 '21 at 09:09
I called onNext from multiple threads. Why shouldn't I call theme from multiple threads? What is the problem of that? If I call theme in one thread, they're printed in order. – Hadi Jul 02 '21 at 09:26
1

`onNext` is not thread safe. Use `create().toSerialized()` to make it thread safe. – akarnokd Jul 02 '21 at 09:30
Thanks. `toSerialized()` solved the problem. – Hadi Jul 02 '21 at 09:35
Sorry but do you know why my second example don't print anything? What is the problem of that?. As I understood `subscribeOn` has no effect on `PublishSubject` because `PublishSubject` has no subscription `side-effect`. But what causes nothing to be printed in my second example? And why outpus are printed if I remove `subscribeOn` from that? – Hadi Jul 02 '21 at 09:47
1

I answered that too. – akarnokd Jul 02 '21 at 09:49
So you mean if I wait a few seconds before calling onNext methods in my second example, messages will be printed, right? – Hadi Jul 02 '21 at 09:54
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/234439/discussion-between-hadi-and-akarnokd). – Hadi Jul 02 '21 at 09:59

Why doesn't subscribeOn effect on PublishSubject in Rxjava?

1 Answers1