Reactive Programming using RxScala

Question

I have a Observable that connects to a service via a Socket protocol. The connection to the socket happens through a client library. The client library that I use has java.util.Observer with with I can register for events being pushed into it

final class MyObservable extends Observable[MyEvent] {

  def subscribe(subscriber: Subscriber[MyEvent]) = {
    // connect to the Socket (Step: 1)
    // get the responses that are pushed (Step: 2)
    // transform them into MyEvent type (Step: 3)
  }
}

I have two open questions that I do not understand.

How can I get the result of Step: 3 in my Subscriber?

Every time when I get a MyEvent, with a subscriber like below, I see that there is a new connection being created. Eventually Step 1, Step 2 and Step 3 are run for each incoming event.

val myObservable = new MyObservale()
myObservable.subscribe()

Where is your Subscriber? Could you use myObservable.subscribe(mySubscriber)? — zsxwing
– zsxwing, Commented Jul 26, 2015 at 14:44
But how to push the result of Step: 3 in the subscribe method of MyObservable class to the outside world? — joesan
– joesan, Commented Jul 27, 2015 at 13:42
Is there a RxScala library that is worthy to consider? How is the NetFlix's RxScala? Has it got back pressure enabling? — joesan
– joesan, Commented Jul 27, 2015 at 14:32
I'm using the monifu library which is an implementation inspired by the Rx.NET. The monifu library has got a lot of goodies which at the moment for me is a bit hard to grasp! — joesan
– joesan, Commented Jul 27, 2015 at 15:54

Brandon · Accepted Answer · 2015-07-29 15:11:32Z

2

Unless I'm misunderstanding your question, you just call onNext:

def subscribe(subscriber: Subscriber[MyEvent]) = {
  // connect to the Socket (Step: 1)
  // get the responses that are pushed (Step: 2)
  // transform them into MyEvent type (Step: 3)

  // finally notify the subscriber:
  subscriber.onNext(myEventFromStep3)
}

and code that subscribes would do something like:

myObservable.subscribe(onNext = println(_))

answered Jul 29, 2015 at 15:11

Brandon

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