Why Rx.Observable.of(true) starts to emit

Question

I have following code:

console.log('start');

//emit value every second
const message = Rx.Observable.interval(1000);
//emit value as soon as subscribed
const trueObs = () => Rx.Observable.of(true);
// start emitting delayed values as soon as trueObs emits
const delayWhenExample = message.delayWhen(trueObs);
//log values start logging after one second
//ex. output: 0...1...2...3
const subscribe = delayWhenExample.subscribe(val => console.log(val));

<script src="https://npmcdn.com/@reactivex/rxjs@5.0.0-beta.8/dist/global/Rx.umd.js"></script>

Why Rx.Observable.of(true) starts to emit value without subscribe the observable?
I understand the concept to RxJS as lazy evaluation, it does not emit values, until I am asking it.

I'm not sure I understand the question. You are calling `subscribe` on the last line. — Peter Hall, Dec 25 '16 at 12:47
But I do subscribe on `message` observable not on `delayForFiveSeconds` — softshipper, Dec 25 '16 at 13:11

score 1 · Accepted Answer · answered Dec 25 '16 at 13:29

Your variable delayWhenExample is actually a SubscriptionDelayObservable observable, which was created by calling delayWhen(). It keeps track of the original message observable and the observable that you passed to it as an argument (delayForFiveSeconds). When you call subscribe(), it will also subscribe behind the scenes to both of these observables, which it needs to calculate its values.

This is really the whole idea of lazy evaluation. You subscribe to the observable that you need, and all of the other observables that it depends on will be subscribed to automatically - but only when necessary. Unsubscribing is also done automatically, which is very convenient when combining data from a lot of different sources.

Thanks so much for helping. – softshipper Dec 25 '16 at 16:11 — softshipper, Dec 25 '16 at 16:11

score 1 · Answer 2 · edited May 23 '17 at 12:00

I am not too familiar with Rxjs v5 but here is probably what is happening :

you subscribe to delayByExample, which is message.delayWhen(delayForFiveSeconds)
this in turns leads to subscription to delayForFiveSeconds as probably the delayWhen operator subscribes to both its operands i.e. delayWhen(message, delayForFiveSeconds) leads to subscription to both message and delayForFiveSeconds and then the value read from message is passed forward when a value from delayForFiveSeconds is received. But both subscriptions occurs at the time of subscription to delayWhen.
last thing is that Rx.Observable.of(true) will emit synchronously on subscription
so, delayWhenExample.subscribe(val => console.log(val)) will immediately trigger the emission of true

If you want to understand more precisely the chain of subscription taking place, you can refer to the following SO answer which comes with illustrated dataflows :

Hot and Cold observables : are there 'hot' and 'cold' operators?

Why Rx.Observable.of(true) starts to emit

2 Answers2