Is unsubscribe() required in operator teardown?

[palamccc]

As per official operator guide https://rxjs.dev/guide/operators#creating-new-operators-from-scratch

function delay<T>(delayInMillis: number) {
  return (observable: Observable<T>) =>
    new Observable<T>((subscriber) => {
      const subscription = observable.subscribe({
        next(value) { ... },
        error(err) { ... },
        complete() { ... },
      });
 
      // Return the finalization logic. This will be invoked when
      // the result errors, completes, or is unsubscribed.
      return () => {
        subscription.unsubscribe(); // <-----------------------------
        // Clean up our timers.
        for (const timerID [of](https://rxjs.dev/api/index/function/of) allTimerIDs) {
          clearTimeout(timerID);
        }
      };
    });
}

subscription.unsubscribe() is called in the teardown logic. Do you think it's really required? I checked a few RxJS operators code in the RxJS repo; they are not doing it.

[palamccc]

One of the operator, where the subscription is not unsusubscribed.

rxjs/packages/rxjs/src/internal/operators/bufferWhen.ts

Lines 81 to 95 in f9f07f1

	source.subscribe(
	operate({
	destination: subscriber,
	// Add every new value to the current buffer.
	next: (value) => buffer?.push(value),
	// When we complete, emit the buffer if we have one,
	// then complete the result.
	complete: () => {
	buffer && subscriber.next(buffer);
	subscriber.complete();
	},
	// Release memory on finalization
	finalize: () => (buffer = closingSubscriber = null!),
	})
	);