My goal is to understand how CompletableFuture works.
My expected result: If I do CompletableFuture.runAsync().thenRun().thenRunAsync(). The thread will be executed in sequence runAsync() -> thenRun() -> thenRunAsync().
My actual result: The sequence is race condition. Sometimes:
runAsync -> thenRunAsync+e -> ...
runAsync -> thenRun -> ...
Reference from SO
public class RunExample5 {
public static void main(String[] args) {
ExecutorService e = Executors.newSingleThreadExecutor(r -> new Thread(r, "sole thread"));
CompletableFuture<?> f = CompletableFuture.runAsync(() -> {
System.out.println("runAsync:\t" + Thread.currentThread());
LockSupport.parkNanos((int) 1e9);
}, e);
f.thenRun(() -> System.out.println("thenRun:\t" + Thread.currentThread()));
f.thenRunAsync(() -> System.out.println("thenRunAsync:\t" + Thread.currentThread()));
f.thenRunAsync(() -> System.out.println("thenRunAsync+e:\t" + Thread.currentThread()),
e);
LockSupport.parkNanos((int) 2e9);
e.shutdown();
}
}
You need
f.thenRun(() -> System.out.println("thenRun:\t" + Thread.currentThread()))
.thenRunAsync(() -> System.out.println("thenRunAsync:\t" + Thread.currentThread()))
.thenRunAsync(() -> System.out.println("thenRunAsync+e:\t" + Thread.currentThread()), e);
The interface to CompletableFuture doesn't work quite the way you're imagining. f itself doesn't keep track of every call to thenRun or thenRunAsync and run them in order; instead, it treats everything from thenRun or thenRunAsync as simultaneously runnable as soon as the main work completes. If you want to chain more complicated sequences of work, you need to use the return value of thenRun or thenRunAsync -- a CompletionStage object -- and call thenRunAsync on that.
Below is my code snippet.
I know you are not supposed to block cachedFlowable like this, but this is just an example.
It gets stuck at the blockingGet line.
If I replace singleOrError with singleElement, the code will still get stuck. If I replace singleOrError with firstElement, the code will no longer get stuck.
Can someone please explain to me why this is the case?
public static void main(String[] args) {
final Flowable<Integer> cachedFlowable = Flowable.just(1).cache();
cachedFlowable
.doOnNext(i -> {
System.out.println("doOnNext " + i);
final Integer j = cachedFlowable.singleOrError().blockingGet();
System.out.println("after blockingGet " + j);
})
.blockingSubscribe();
}
The reason it deadlocks with singleX operator is that such operators wait for a possible 2nd item emission but since you are blocking them, any second item or completion from the main source can't get executed. With firstX they only care about the very first item thus unblock almost immediately which allows the source to complete.
So yes, you should not use blocking methods in flows like that but instead use flatMap or concatMap to do a per item subflow:
var cache = Flowable.just(1).cache();
cache
.doOnNext(i -> System.out.println("doOnNext " + i))
.concatMapSingle(item -> cache.firstOrError())
.doOnNext(j -> System.out.println("after " + j))
.blockingSubscribe();
I'm new to RX java and I've been experimenting with observeOn and subscribeOn methods. I've read that the difference between them is that they affect the whole chain(subscribeOn) or only the part of the chain after setting a scheduler(observeOn). So why the code below executes fine (prints the current thread):
Observable obs = Observable.from(Arrays.asList("element", "2nd element"));
obs.observeOn(Schedulers.newThread())
.map(x -> x.toString().toUpperCase())
.subscribe(x -> System.out.println("NT:" + Thread.currentThread().getName() + x));
while this code doesn't print anything:
Observable obs = Observable.from(Arrays.asList("element", "2nd element"));
obs.subscribeOn(Schedulers.newThread())
.map(x -> x.toString().toUpperCase())
.subscribe(x -> System.out.println("NT:" + Thread.currentThread().getName() + x));
Are you sure that this code doesn't print anything?
I tried this code:
Observable obs = Observable.from(Arrays.asList("element", "2nd element"));
obs.subscribeOn(Schedulers.newThread())
.map(x -> x.toString().toUpperCase())
.subscribe(x -> System.out.println("NT:" + Thread.currentThread().getName() + x));
Thread.sleep(5000);
Output:
NT:RxNewThreadScheduler-1ELEMENT
NT:RxNewThreadScheduler-12ND ELEMENT
Maybe you forgot to sleep or do some another work to make application wait for completion of new RxJava threads?
Basically, an observable (more specifically, a cold observable) is a delayed calculation. So, .subscribeOn() defined on which scheduler/thread this calculation will be done, and .observeOn() defines on which scheduler/thread you will receive its result. For instance, network request should run on Schedulers.io which you specify by adding .subscribeOn(Schedulers.io) and you want to display results on main thread by specifying .observeOn(AndroidShedulers.main)
I am just learning Rx-java and Rxandroid2 and I am just confused what is the major difference between in SubscribeOn and ObserveOn.
SubscribeOn specify the Scheduler on which an Observable will operate.
ObserveOn specify the Scheduler on which an observer will observe this Observable.
So basically SubscribeOn is mostly subscribed (executed) on a background thread ( you do not want to block the UI thread while waiting for the observable) and also in ObserveOn you want to observe the result on a main thread...
If you are familiar with AsyncTask then SubscribeOn is similar to doInBackground method and ObserveOn to onPostExecute...
In case you find the above answer full of jargons:
tl;dr
Observable.just("Some string")
.map(str -> str.length())
.observeOn(Schedulers.computation())
.map(length -> 2 * length)
.observeOn(AndroidSchedulers.mainThread())
.subscribeOn(Schedulers.io())
.subscribe(---)
Observe an observable... perform the map function in an IO thread (since we are "subscribingOn" that thread)... now switch to a Computation Thread and perform map(length -> 2 * length) function... and finally make sure you Observe the output on (observeOn()) Main thread.
Anyway,
observeOn() simply changes the thread of all operators further Downstream. People usually have this misconception that observeOn also acts as upstream, but it doesn't.
The below example will explain it better...
Observable.just("Some string") // UI
.map(str -> str.length()) // UI
.observeOn(Schedulers.computation()) // Changing the thread
.map(length -> 2 * length) // Computation
.subscribe(---) // Computation
subscribeOn() only influences the thread which is going to be used when Observable is going to get subscribed to and it will stay on it downstream.
Observable.just("Some String") // Computation
.map(str -> str.length()) // Computation
.map(length -> 2 * length) // Computation
.subscribeOn(Schedulers.computation()) // -- changing the thread
.subscribe(number -> Log.d("", "Number " + number)); // Computation
Position does not matter (subscribeOn())
Why?
Because it affects only the time of subscription.
Methods that obey the contact with subscribeOn
-> Basic example : Observable.create
All the work specified inside the create body will run on the thread specified in subscribeOn.
Another example: Observable.just,Observable.from or Observable.range
Note: All those methods accept values, so do not use blocking methods to create those values, as subscribeOn won't affect it.
If you want to use blocking functions, use
Observable.defer(() -> Obervable.just(blockingMenthod())));
Important Fact:
subscribeOn does not work with Subjects
Multiple subscribeOn:
If there are multiple instances of subscribeOn in the stream, only the first one has a practical effect.
Subscribe & subscribeOn
People think that subscribeOn has something to do with Observable.subscribe, but it doesn't have anything special to do with it.
It only affects the subscription phase.
Source : Tomek Polański (Medium)
Summary
Use observeOn to set threads for callbacks "further down the stream (below it)", such as code blocks inside doOnNext or map.
Use subscribeOn to set threads for initializations "upstream (above it)", such as doOnSubscribe, Observable.just or Observable.create.
Both methods can be called multiple times, with each call overwriting previous ones. Position matters.
Let's walk through this topic with an example: we want to find the length of the string "user1032613". This is not an easy task for computers, so it's only natural that we perform the intense calculation in a background thread, to avoid freezing the app.
observeOn
We can call observeOn as many times as we like, and it controls which thread all callbacks below it will run. It's easy to use, and works just as you'd expect.
For example, we will show a progress bar on the main UI thread, then do intensive/blocking operations in another thread, then come back to the main UI thread to update the result:
Observable.just("user1032613")
.observeOn(mainThread) // set thread for operation 1
.doOnNext {
/* operation 1 */
print("display progress bar")
progressBar.visibility = View.VISIBLE
}
.observeOn(backThread) // set thread for operation 2 and 3
.map {
/* operation 2 */
print("calculating")
Thread.sleep(5000)
it.length
}
.doOnNext {
/* operation 3 */
print("finished calculating")
}
.observeOn(mainThread) // set thread for operation 4
.doOnNext {
/* operation 4 */
print("hide progress bar and display result")
progressBar.visibility = View.GONE
resultTextView.text = "There're $it characters!"
}
.subscribe()
In the above example, /* operation 1 */ is ran in the mainThread because we set it using observeOn(mainThread) on the line right above it; then we switch to backThread by calling observeOn again, so /* operation 2 */ will run there. Because we didn't change it before chaining /* operation 3 */, it will run in the back thread as well, just like /* operation 2 */; finally we call observeOn(mainThread) again, to make sure /* operation 4 */ updates the UI from the main thread.
subscribeOn
So we've learned observeOn sets threads for subsequent callbacks. What else are we missing? Well, the Observable itself, and its methods such as just(), create(), subscribe() and so on, are also code that needs to be executed. This is how objects are passed along the stream. We use subscribeOn to set threads for code related to Observable itself.
If we remove all the callbacks (controlled by observeOn discussed earlier), we are left with the "skeleton code" that will, by default, run on whichever thread the code is written in (probably main thread):
Observable.just("user1032613")
.observeOn(mainThread)
.doOnNext {
}
.observeOn(backThread)
.map {
}
.doOnNext {
}
.observeOn(mainThread)
.doOnNext {
}
.subscribe()
If we aren't happy about this empty skeleton code running on main thread, we can use subscribeOn to change it. For example, maybe the first line Observable.just("user1032613") isn't as simple as creating a stream from my user name - maybe it's a string from the Internet, or perhaps you are using doOnSubscribe for some other intensive operations. In that case, you can call subscribeOn(backThread) to put some of the code in another thread.
Where to put subscribeOn
At the time of writing this answer, there are some misconceptions saying "only call it once", "position does not matter", and "if you call it multiple times, only the first time counts". After lots of researches and experiments, it turns out subscribeOn can be usefully called multiple times.
Because Observable uses Builder Pattern (fancy name for "chaining methods one after another"), subscribeOn is applied in reverse order. Therefore, this method sets the thread for code above it, exactly the opposite of observeOn.
We can experiment this using doOnSubscribe method. This method is triggered on the subscription event, and it runs on the thread set by subscribeOn:
Observable.just("user1032613")
.doOnSubscribe {
print("#3 running on main thread")
}
.subscribeOn(mainThread) // set thread for #3 and just()
.doOnNext {
}
.map {
}
.doOnSubscribe {
print("#2 running on back thread")
}
.doOnNext {
}
.subscribeOn(backThread) // set thread for #2 above
.doOnNext {
}
.doOnSubscribe {
print("#1 running on default thread")
}
.subscribe()
It might be easier to follow the logic, if you read the above example from bottom to top, just like how Builder Pattern executes the code.
In this example, the first line Observable.just("user1032613") is run in the same thread as print("#3") because there are no more subscribeOn in-between them. This creates the illusion of "only the first call matters" for people who only care about code inside just() or create(). This quickly falls apart once you start doing more.
Footnote:
Threads and print() functions in the examples are defined, for brevity, as follows:
val mainThread = AndroidSchedulers.mainThread()
val backThread = Schedulers.computation()
private fun print(msg: String) = Log.i("", "${Thread.currentThread().name}: $msg")
If someone finds rx java description hard to understand (as me for example), here is pure java explanation:
subscribeOn()
Observable.just("something")
.subscribeOn(Schedulers.newThread())
.subscribe(...);
Is equivalent of:
Observable observable = Observable.just("something");
new Thread(() -> observable.subscribe(...)).start();
Because Observable emits values on subscribe() and here subscribe() goes in the separate thread, the values are also emitted in the same thread as subscribe(). This is why it works "upstream" (influences the thread for the previous operations) and "downstream".
observeOn()
Observable.just("something")
.observeOn(Schedulers.newThread())
.subscribe(...);
Is equivalent of:
Observable observable = Observable.just("something")
.subscribe(it -> new Thread(() -> ...).start());
Here Observable emits values in the main thread, only the listener method is executed in the separate thread.
When you subscribe to an observable, a flow starts that works its way up to the top of chain and then back down again. The subscribe part is relevant to the upward chaining and the observe part is relevant to the downward chaining.
Once the top of the chain is reached, the subscription phase has essentially completed. Events start to be emitted and the downward chain of maps, filters etc are invoked.
SubscribeOn influences subscription calls above its placement, for example doOnSubscribe.
ObserveOn influences observation calls below its placement, for example, doOnNext, map, flatmap etc.
Both will change the thread that is used to continue the flow either upward or downward.
import io.reactivex.Observable;
import io.reactivex.schedulers.Schedulers;
import java.util.concurrent.CountDownLatch;
public class SubscribeVsObserveOn {
public static void main(String[] args) throws InterruptedException {
System.out.println("Ordinal 0: " + Thread.currentThread().getName());
final CountDownLatch latch = new CountDownLatch(1);
Observable
.just("a regular string.")
.doOnSubscribe(disposable ->
System.out.println("Ordinal 2: " + Thread.currentThread().getName()))
.subscribeOn(Schedulers.newThread())
.observeOn(Schedulers.newThread())
.doOnNext(s ->
System.out.println("Ordinal 3: " + Thread.currentThread().getName()))
.map(s -> s)
.doOnSubscribe(disposable ->
System.out.println("Ordinal 1: " + Thread.currentThread().getName()))
.subscribeOn(Schedulers.newThread())
.observeOn(Schedulers.newThread())
.doOnNext(s ->
System.out.println("Ordinal 4: " + Thread.currentThread().getName()))
.map(s -> s)
.subscribe(s -> latch.countDown());
latch.await();
}
}
Here is the output:
Ordinal 0: main
Ordinal 1: RxNewThreadScheduler-1
Ordinal 2: RxNewThreadScheduler-2
Ordinal 3: RxNewThreadScheduler-3
Ordinal 4: RxNewThreadScheduler-4
This answer is nothing new, I just want to clarify a little bit more.
Let's assume that we have two threads.
val pool1 = Executors.newCachedThreadPool { runnable -> Thread(runnable, "Thread 1") }
val pool2 = Executors.newCachedThreadPool { runnable -> Thread(runnable, "Thread 2") }
As the answers described, observeOn will set Downstream, and subscribeOn will set Upstream. But what if both of them was used? For check this, I added logs line by line.
Observable.just("what if use both")
.doOnSubscribe { Log.d("Thread", "both, doOnSubscribe A " + Thread.currentThread().name) }
.doOnNext { Log.d("Thread", "both, doOnNext A " + Thread.currentThread().name) }
.map {
Log.d("Thread", "both, map A " + Thread.currentThread().name)
it + " A"
}
// observeOn
.observeOn(Schedulers.from(pool1))
.doOnSubscribe { Log.d("Thread", "both, doOnSubscribe B " + Thread.currentThread().name) }
.doOnNext { Log.d("Thread", "both, doOnNext B " + Thread.currentThread().name) }
.map {
Log.d("Thread", "both, map B " + Thread.currentThread().name)
it + " B"
}
// subscribeOn
.subscribeOn(Schedulers.from(pool2))
.doOnSubscribe { Log.d("Thread", "both, doOnSubscribe C " + Thread.currentThread().name) }
.doOnNext { Log.d("Thread", "both, doOnNext C " + Thread.currentThread().name) }
.map {
Log.d("Thread", "both, map C " + Thread.currentThread().name)
it + " C"
}
// observeOn main
.observeOn(AndroidSchedulers.mainThread())
.doOnNext { Log.d("Thread", "main " + Thread.currentThread().name) }
.subscribe(
{ result -> Log.d("Thread", "main subscribe " + Thread.currentThread().name)}
, { error -> {} }
)
The result is:
both, doOnSubscribe C main
both, doOnSubscribe A Thread 2
both, doOnSubscribe B Thread 2
both, doOnNext A Thread 2
both, map A Thread 2
both, doOnNext B Thread 1
both, map B Thread 1
both, doOnNext C Thread 1
both, map C Thread 1
main main
main subscribe main
result: what if use both A B C
As you can see, doOnSubscribe called first, from bottom to top. That means subscribe has priority over other operators, so the first thread which handles first code was Thread 2.
And then other operators was called, line by line. After observeOn, thread was changed to Thread 1. Then, just before calling subscribe, observeOn was called again, for change thread to main thread. (Don't care about AndroidSchedulers, it is just a kind of scheduler)
TL;DR;
First path, subscribeOn called first, from bottom to top.
Second path, observeOn called, from top to bottom, along with other codes.
Behavior was same on both RxJava2 and RxJava3
Using RxJava2, I am trying to accomplish some timed events.
I have a few async events that might complete in less than one second, and I want to be displaying a message to the user for at least one second, before continuing on with the result of the async operation (and maybe use that result later).
The async operation may also timeout, and I would want to display a message and not continue.
I can accomplish this by using an zip() with the timer as the first parameter and the async operation as the second operator, but what do I do with the next 'layer'?
This is the code I have so far, which actually does work, but I feel very dirty creating nested subscriptions (using just() in place of the async operation, and ignoring subscription threads)
mStrings is just a BehaviorSubject<String>.
mStrings.onNext("Waiting 1 second. The result will fire at the Single.timer onComplete");
Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("First"), (t1, t2) -> t2)
.timeout(15, TimeUnit.SECONDS, observer -> {
mStrings.onNext("First timeout fired");
})
.subscribe(s1 -> {
mStrings.onNext("First timer fired and returned " + s1);
Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("Second"), (t1, t2) -> t2)
.timeout(15, TimeUnit.SECONDS, observer -> {
mStrings.onNext("Second timeout fired");
})
.subscribe(s2 -> {
mStrings.onNext("Second timer fired and returned " + s2 + ". Previous was " + s1);
Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("Third"), (t1, t2) -> t2)
.timeout(15, TimeUnit.SECONDS, observer -> {
mStrings.onNext("Third timeout fired");
})
.subscribe(s3 -> {
mStrings.onNext("Third timer fired and returned " + s3 + ". Previous was " + s1 + " and " + s2 );
});
});
});
The result of which is:
17:53:53.219 Waiting 1 second. The result will fire at the Single.timer onComplete
17:53:54.220 First timer fired and returned First
17:53:55.224 Second timer fired and returned Second. Previous was First
17:53:56.224 Third timer fired and returned Third. Previous was First and Second
Am I missing an operator that would make sense in this type of flow? Or some elementary methodology? I know I might be able to work out an alternative solution using multiple subjects, but it seems excessive.
I think I fixed my own issue. flatMap can be used to condense all the subscriptions into one, and each individual timer and timeout will still operate as expected.
mStrings.onNext("Waiting 1 second. The result will fire at the Single.timer onComplete");
Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("First"), (t1, t2) -> t2)
.timeout(2, TimeUnit.SECONDS, observer -> {
mStrings.onNext("First timeout fired");
})
.flatMap(s1 -> {
mStrings.onNext("First timer fired and returned " + s1);
return Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("Second"), (t1, t2) -> t2)
.timeout(15, TimeUnit.SECONDS, observer -> {
mStrings.onNext("Second timeout fired");
})
.flatMap(s2 -> {
mStrings.onNext("Second timer fired and returned " + s2 + ". Previous was " + s1);
return Single.zip(Single.timer(1, TimeUnit.SECONDS), Single.just("Third"), (t1, t2) -> t2)
.timeout(15, TimeUnit.SECONDS, observer -> {
mStrings.onNext("Third timeout fired");
})
.flatMap(s3 -> {
mStrings.onNext("Third timer fired and returned " + s3 + ". Previous was " + s1 + " and " + s2 );
return Single.just("Fourth");
});
});
})
.subscribe(s -> {
// ignored
});
}));
The complex nesting can still be simplified by composing the Observables a bit more logically.