CODEWITHSUNDEEP
androidrx-java
Alex Fu
Alex Fu

Reputation: 5529

Android RxJava, Non Blocking?

It was my understanding that rxjava-android performs operations on a separate thread (when provided the correct Scheduler), leading to non-blocking operations, however a quick and dirty test seems to prove this to be incorrect.

I used the following code snippets and in both, scenarios, the UI was being blocked...

Snippet 1

Observable observable = Observable.create(new Observable.OnSubscribe<Object>() {
  @Override
  public void call(Subscriber<? super Object> subscriber) {
    int i = 0;
    while (i == 0) {}
    subscriber.onCompleted();
  }
});
observable.subscribeOn(Schedulers.newThread());
observable.observeOn(AndroidSchedulers.mainThread());
observable.subscribe();

Snippet 2

Observable observable = Observable.create(new Observable.OnSubscribe<Object>() {
  @Override
  public void call(Subscriber<? super Object> subscriber) {
    SystemClock.sleep(5000);
    subscriber.onCompleted();
  }
});
observable.subscribeOn(Schedulers.newThread());
observable.observeOn(AndroidSchedulers.mainThread());
observable.subscribe();

Am I missing something here?

Upvotes: 9

Views: 5183

Answers (1)

zsxwing
zsxwing

Reputation: 20816

The mistake is that you use the wrong Observable. The correct code should be:

Observable observable = Observable.create(new Observable.OnSubscribe<Object>() {
  @Override
  public void call(Subscriber<? super Object> subscriber) {
    int i = 0;
    while (i == 0) {}
    subscriber.onCompleted();
  }
});
observable.subscribeOn(Schedulers.newThread()).observeOn(AndroidSchedulers.mainThread()).subscribe();

Both subscribeOn and observeOn return a new Observable which implements their functions. But the original Observable is not modified. Actually, every operator will always create a new Observable without modifying the original one.

Upvotes: 29

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