In the following code:-
RxJS.Observable.of(1,2).first().subscribe((x) => console.log(x););
is it necessary to unsubscribe given the operator first()?
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3 Answers
46
No. It unsubscribes automatically after calling first(). The current syntax is observable.pipe(first()).subscribe(func); for RxJS 6.
The documentation states:
If called with no arguments,
firstemits the first value of the source Observable, then completes.
Max
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Xavier Reyes Ochoa
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Think you are missing a `)` :) – El Ronnoco Jan 28 '19 at 11:57
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8what happens when the source observable never emits a value, does that mean that the Subscription stays there and possibly create a memory leak ? – Gagan Sep 19 '19 at 16:05
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@Gagan Yes, it can create a memory leak. – Max Nov 10 '20 at 08:18
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1So how to add unsubscribing in this case? – Tony Nov 23 '20 at 12:34
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takeUntil(this.destroyed$) Where destroyed$ gets set on ngOnDestroy() – Micaël Félix Aug 28 '23 at 08:02
6
For the example provided, you dont need to unsubscribe, and neither you need to call first, as Observable.of(1) actually completes after emitting its first (and last) value.
c69
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1Changed to 1,2 but your answer still is correct since you mentioned it Completes. – bhantol Apr 06 '18 at 11:49
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first() will complete after the first item is emitted from the observable.
Also subscribe() takes three arguments, the last one being the complete callback. Running the following code will output 1 followed by 'done'.
Rx.Observable.of(1)
.subscribe(
(x) => console.log(x), // next
(x) => console.error(x), // error
() => console.log('done') // done
)
Johan.C
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