为什么subscribeOn在RXJava中对PublishSubject不起作用？

Question

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)
}

我试图在不同的IO线程中运行map块和subscribe的onnext块。但是输出是这样的：

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

如您所见，调用subscribeon似乎对publissubject流没有影响，thread-0、thread-1和thread-2引用调用onnext方法的线程。

另外，请考虑下面的代码：

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)
}

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

这些代码的问题是什么？谢了。

Answer 1

因为subscribeon只影响源的订阅副作用。如果源在观察者订阅时就开始发出事件，则会产生这样的副作用：

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

PublishSubject没有订阅副作用，因为它只将信号从其onxxx方法中继到观察者的onxxx方法。

但是，subscribeon具有时间效应，因为它会延迟对源的实际订阅，因此在publissubject的情况下，它可能无法及时看到注册的观察者，因为其他线程调用了它的onxxx方法。

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}")
        })