CODEWITHSUNDEEP
kotlinrx-javarx-java2
ericn
ericn

Reputation: 13103

subscribeOn(Schedulers.io()) not working but subscribeOn(Schedulers.trampoline()) does

fun main() {
    Single.just("faefw")
        .subscribeOn(Schedulers.trampoline())
        .map {
            println("map in thread ${Thread.currentThread().name}")
        }
        .flatMap {
            return@flatMap Single.just(231).map {
                println("map in thread ${Thread.currentThread().name}")
                return@map it
            }
        }
        .subscribe({
            println(it)
        }, {
            println(it)
        })
}

prints out:

map in thread main
map in thread main
231

But

fun main() {
    Single.just("faefw")
        .subscribeOn(Schedulers.io())
        .map {
            println("map in thread ${Thread.currentThread().name}")
        }
        .flatMap {
            return@flatMap Single.just(231).map {
                println("map in thread ${Thread.currentThread().name}")
                return@map it
            }
        }
        .subscribe({
            println(it)
        }, {
            println(it)
        })
}

outputs nothing!

subscribeOn(Schedulers.computation()) and subscribeOn(Schedulers.newThread()) do not work either

Why?

Upvotes: 0

Views: 582

Answers (1)

Aarjav
Aarjav

Reputation: 1374

The program is exiting before the code has time to execute. Try with a sleep or put in a print statement after subscribe

fun main() {
    Single.just("faefw")
        .subscribeOn(Schedulers.io())
        .map {
            println("map in thread ${Thread.currentThread().name}")
        }
        .flatMap {
            return@flatMap Single.just(231).map {
                println("map in thread ${Thread.currentThread().name}")
                return@map it
            }
        }
        .subscribe({
            println(it)
        }, {
            println(it)
        })
    println("just subscribed in last statement!")
    Thread.sleep(500)
    println("exiting main thread...")
}

with trampoline the output is

map in thread main
map in thread main
231
just subscribed in last statement!
exiting main thread...

with other schedulers which do the "work" in different threads (like .io() and .computation()) will have output like this:

just subscribed in last statement!
map in thread RxComputationThreadPool-1
map in thread RxComputationThreadPool-1
231
exiting main thread...

PS. The io() scheduler you can think of as unbounded thread pool, computation() as a bounded thread pool.

Upvotes: 1

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