What is the difference between Rx.Observable subscribe and forEach


Rxjs Problem Overview

After creating an Observable like so

var source = Rx.Observable.create(function(observer) {...});

What is the difference between subscribe

source.subscribe(function(x) {});

and forEach

source.forEach(function(x) {});

Rxjs Solutions

Solution 1 - Rxjs

In the ES7 spec, which RxJS 5.0 follows (but RxJS 4.0 does not), the two are NOT the same.


public subscribe(observerOrNext: Observer | Function, error: Function, complete: Function): Subscription

Observable.subscribe is where you will do most of your true Observable handling. It returns a subscription token, which you can use to cancel your subscription. This is important when you do not know the duration of the events/sequence you have subscribed to, or if you may need to stop listening before a known duration.


public forEach(next: Function, PromiseCtor?: PromiseConstructor): Promise

Observable.forEach returns a promise that will either resolve or reject when the Observable completes or errors. It is intended to clarify situations where you are processing an observable sequence of bounded/finite duration in a more 'synchronous' manner, such as collating all the incoming values and then presenting once, by handling the promise.

Effectively, you can act on each value, as well as error and completion events either way. So the most significant functional difference is the inability to cancel a promise.

Solution 2 - Rxjs

I just review the latest code available, technically the code of foreach is actually calling subscribe in RxScala, RxJS, and RxJava. It doesn't seems a big different. They now have a return type allowing user to have an way for stopping a subscription or similar.

When I work on the RxJava earlier version, the subscribe has a subscription return, and forEach is just a void. Which you may see some different answer due to the changes.

 * Subscribes to the [[Observable]] and receives notifications for each element.
 * Alias to `subscribe(T => Unit)`.
 * $noDefaultScheduler
 * @param onNext function to execute for each item.
 * @throws java.lang.IllegalArgumentException if `onNext` is null
 * @throws rx.exceptions.OnErrorNotImplementedException if the [[Observable]] tries to call `onError`
 * @since 0.19
 * @see <a href="http://reactivex.io/documentation/operators/subscribe.html">ReactiveX operators documentation: Subscribe</a>
def foreach(onNext: T => Unit): Unit = {

def subscribe(onNext: T => Unit): Subscription = {

 *  Subscribes an o to the observable sequence.
 *  @param {Mixed} [oOrOnNext] The object that is to receive notifications or an action to invoke for each element in the observable sequence.
 *  @param {Function} [onError] Action to invoke upon exceptional termination of the observable sequence.
 *  @param {Function} [onCompleted] Action to invoke upon graceful termination of the observable sequence.
 *  @returns {Disposable} A disposable handling the subscriptions and unsubscriptions.
observableProto.subscribe = observableProto.forEach = function (oOrOnNext, onError, onCompleted) {
  return this._subscribe(typeof oOrOnNext === 'object' ?
    oOrOnNext :
    observerCreate(oOrOnNext, onError, onCompleted));

 * Subscribes to the {@link Observable} and receives notifications for each element.
 * <p>
 * Alias to {@link #subscribe(Action1)}
 * <dl>
 *  <dt><b>Scheduler:</b></dt>
 *  <dd>{@code forEach} does not operate by default on a particular {@link Scheduler}.</dd>
 * </dl>
 * @param onNext
 *            {@link Action1} to execute for each item.
 * @throws IllegalArgumentException
 *             if {@code onNext} is null
 * @throws OnErrorNotImplementedException
 *             if the Observable calls {@code onError}
 * @see <a href="http://reactivex.io/documentation/operators/subscribe.html">ReactiveX operators documentation: Subscribe</a>
public final void forEach(final Action1<? super T> onNext) {

public final Disposable forEach(Consumer<? super T> onNext) {
    return subscribe(onNext);


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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionKevin Le - KhnleView Question on Stackoverflow
Solution 1 - RxjsshannonView Answer on Stackoverflow
Solution 2 - Rxjsphdfong - Kenneth FongView Answer on Stackoverflow