Rxjava: Subscribe on the specific thread - multithreading

Question

I'm a newbie in Rxjava.
I have the following code:
System.out.println("1: " + Thread.currentThread().getId());
Observable.create(new rx.Observable.OnSubscribe<String>() {
#Override
public void call(Subscriber<? super String> subcriber) {
System.out.println("2: " + Thread.currentThread().getId());
// query database
String result = ....
subcriber.onNext(result);
}
}).subscribeOn(Schedulers.newThread()).subscribe(countResult -> {
System.out.println("3: " + Thread.currentThread().getId());
});
For example, the output will be:
1: 50
2: 100
3: 100
I want subscribers run on the thread that has id 50. How can I do that?

Answer 1

I think that there are two cases. Either you need it to run on the UI thread, or because of synchronisation. As I know you can not call a function on a specific thread, because when the method is called it is bound to the context of the thread, so it is impossible to call a method from a thread to another thread. Your problem is that the method in subscriber is called from Schedulers.newThread(). I also found this github issue about Schedulers.currentThread(). What you need is to notify the caller thread when the observer gets called.
Also you can use akka, it is way simpler to write concurrent code with it.
Sorry for my bad grammar.

Answer 2

From the docs:
By default, an Observable and the chain of operators that you apply to
it will do its work, and will notify its observers, on the same thread
on which its Subscribe method is called. The SubscribeOn operator
changes this behavior by specifying a different Scheduler on which the
Observable should operate. The ObserveOn operator specifies a
different Scheduler that the Observable will use to send notifications
to its observers.
So you can just use subscribe instead of subscribeOn to observe your collection on the same thread it was created, something like this:
Observable.create(new rx.Observable.OnSubscribe<String>() {
#Override
public void call(Subscriber<? super String> subcriber) {
System.out.println("2: " + Thread.currentThread().getId());
// query database
String result = ....
subcriber.onNext(result);
}
}).subscribe(countResult -> {
System.out.println("3: " + Thread.currentThread().getId());
});
UPDATE:
If your application is an Android application, you can use subscribe on a background thread as you do and pass the results to the main thread using Handler messages.
If your application is a Java application I may suggest using wait() and notify() mechanism or consider using frameworks such as EventBus or akka for more complex scenarios.

Answer 3

With RxJava 1.0.15, you can apply toBlocking() before subscribe and everything will run on the thread that created the entire sequence of operators.

Answer 4

So subscribeOn denotes what thread the Observable will start emitting items on, observeOn "switches" to a thread for the rest of the observable chain. Put a observeOn(schedulers.CurrentThread()) right before the subscribe and you'll be in the thread that this observable is created in rather than the thread it is executed in. Here's a resource that explains rxjava threading really well. http://www.grahamlea.com/2014/07/rxjava-threading-examples/

Answer 5

I believe #akarnokd is right. You either run without the .subscribeOn(Schedulers.newThread()) so that it happens synchronously or use toBlocking() just before the subscribe. Alternatively if you just want everything to happen on the same thread but it doesn't have to be the current thread then you can do this:
Observable
.defer(() -> Observable.create(...))
.subscribeOn(Schedulers.newThread())
.subscribe(subscriber);
defer ensures that the call to create happens on the subscription thread.