Completion order of CompletableFuture - java

I'm curious to understand the completion order when a CompletableFuture has multiple dependents. I would have expected the dependents to complete in the order they were added, but that doesn't seem to be the case.
In particular, this behaviour is surprising:
import java.util.concurrent.CompletableFuture;
public class Main {
public static void main(String[] args) {
{
var x = new CompletableFuture<Void>();
x.thenRun(() -> System.out.println(1));
x.thenRun(() -> System.out.println(2));
x.thenRun(() -> System.out.println(3));
x.complete(null);
}
{
var x = new CompletableFuture<Void>();
var y = x.copy();
y.thenRun(() -> System.out.println(1));
y.thenRun(() -> System.out.println(2));
y.thenRun(() -> System.out.println(3));
x.complete(null);
}
}
}
...results in the the following output...
3
2
1
1
2
3

All 3 child futures generated by 'x.thenRun()' expression run in parallel, so you cannot expect any particular order of printed numbers.
The same for 'y.thenRun()'.
The expressed order is an implementation feature, can change in the future, and is not supported by the specification.

So you have not chained those thenRun, as such there are zero guarantees on the order, but you expect some order? Sorry, that's not how it works.
The documentation does not say anything about order of execution in such cases, so whatever you see now in the output:
is not a guarantee
can change from run to run (and from java version to version)
It is a little bit hard to prove via code with your current set-up, but just change it to :
var x = CompletableFuture.runAsync(() -> {
LockSupport.parkNanos(TimeUnit.SECONDS.toNanos(1));
});
x.thenRun(() -> System.out.println(1));
x.thenRun(() -> System.out.println(2));
x.thenRun(() -> System.out.println(3));
x.join();
And run this let's say 20 times, I bet that at least once, you will not see 3, 2, 1, but something different.

Related

RxJava retryWhen (exponential back-off) not working

So I know this has been asked many times before, but I have tried many things and nothing seems to work.
Let's start with these blogs/articles/code:
https://blog.danlew.net/2016/01/25/rxjavas-repeatwhen-and-retrywhen-explained/
https://jimbaca.com/rxjava-retrywhen/
http://blog.inching.org/RxJava/2016-12-12-rx-java-error-handling.html
https://pamartinezandres.com/rxjava-2-exponential-backoff-retry-only-when-internet-is-available-5a46188ab175
https://gist.github.com/wotomas/35006d156a16345349a2e4c8e159e122
And many others.
In a nutshell all of them describe how you can use retryWhen to implement exponential back-off. Something like this:
source
.retryWhen(
errors -> {
return errors
.zipWith(Observable.range(1, 3), (n, i) -> i)
.flatMap(
retryCount -> {
System.out.println("retry count " + retryCount);
return Observable.timer((long) Math.pow(1, retryCount), SECONDS);
});
})
Even the documentation in the library agrees with it:
https://github.com/ReactiveX/RxJava/blob/3.x/src/main/java/io/reactivex/rxjava3/core/Observable.java#L11919.
However, I've tried this and some pretty similar variations, not worthy to describe here, and nothing seems to work. There's a way in that the examples works and is using blocking subscribers but I want to avoid blocking threads.
So if to the previous observable we apply a blocking subscriber like this:
.blockingForEach(System.out::println);
It works as expected. But as that's not the idea. If we try:
.subscribe(
x -> System.out.println("onNext: " + x),
Throwable::printStackTrace,
() -> System.out.println("onComplete"));
The flow runs only once, thus not what I want to achieve.
Does that mean it cannot be used as I'm trying to? From the documentation it doesn't seem to be a problem trying to accomplish my requirement.
Any idea what am I missing?
TIA.
Edit: There are 2 ways I'm testing this:
A test method (using testng):
Observable<Integer> source =
Observable.just("test")
.map(
x -> {
System.out.println("trying again");
return Integer.parseInt(x);
});
source
.retryWhen(
errors -> {
return errors
.zipWith(Observable.range(1, 3), (n, i) -> i)
.flatMap(
retryCount -> {
return Observable.timer((long) Math.pow(1, retryCount), SECONDS);
});
})
.subscribe(...);
From a Kafka consumer (using Spring boot):
This is only the subscription to the observer, but the retries logic is what I described earlier in the post.
#KafkaListener(topics = "${kafka.config.topic}")
public void receive(String payload) {
log.info("received payload='{}'", payload);
service
.updateMessage(payload)
.subscribe(...)
.dispose();
}
The main issue of your code is that Observable.timer is by default operating on the computation scheduler. This adds extra effort when trying to verify the behaviour within a test.
Here is some unit testing code that verifies that your retry code is actually retrying.
It adds a counter, just so we can easily check how many calls have happened.
It uses the TestScheduler instead of the computation scheduler so that we can pretend moving in time through advanceTimeBy.
TestScheduler testScheduler = new TestScheduler();
AtomicInteger counter = new AtomicInteger();
Observable<Integer> source =
Observable.just("test")
.map(
x -> {
System.out.println("trying again");
counter.getAndIncrement();
return Integer.parseInt(x);
});
TestObserver<Integer> testObserver = source
.retryWhen(
errors -> {
return errors
.zipWith(Observable.range(1, 3), (n, i) -> i)
.flatMap(
retryCount -> {
return Observable.timer((long) Math.pow(1, retryCount), SECONDS, testScheduler);
});
})
.test();
assertEquals(1, counter.get());
testScheduler.advanceTimeBy(1, SECONDS);
assertEquals(2, counter.get());
testScheduler.advanceTimeBy(1, SECONDS);
assertEquals(3, counter.get());
testScheduler.advanceTimeBy(1, SECONDS);
assertEquals(4, counter.get());
testObserver.assertComplete();

Limit for `onErrorContinue(...)` in Flux?

I have a (possibly infinite) Flux source that is supposed to first store each message (e.g. into a database) and then asynchronously forward the messages (e.g. using Spring WebClient).
The forward(s) in case of failure are supposed to log an error, without completing the source Flux.
I however realized that forward(s) wihtin the flow (flatMap(...)) block execution of the source Flux after exactly 256 messages that cause exceptions (e.g. reactor.retry.RetryExhaustedException).
Representative example that fails in the assert since only 256 messages are processed:
#Test
#SneakyThrows
public void sourceBlockAfter256Exceptions() {
int numberOfRequests = 500;
Set<Integer> sink = new HashSet<>();
Flux
.fromStream(IntStream.range(0, numberOfRequests).boxed())
.map(sink::add)
.flatMap(i -> Mono
// normally the forwards are contained here e.g. by means of Mono.when(...).thenReturn(...).retryWhen(...):
.error(new Exception("any"))
)
.onErrorContinue((throwable, o) -> log.error("Error", throwable))
.subscribe();
Thread.sleep(3000);
Assertions.assertEquals(numberOfRequests, sink.size());
}
Doing the forward within the subscribe(...) doesn't block the source Flux but that's certainly no solution, since I don't possibly want to lose messages.
Questions:
What has happened here? (probably related to some state stored in just one bit)
How can I do this correctly?
EDIT:
According to the discussion below I've constructed an example that uses FluxMessageChannel (which up to my understanding is made for infinite streams and definitly not expected to block after 256 Errors) and has exactly the same behaviour:
#Test
#SneakyThrows
public void maxConnectionWithChannelTest() {
int numberOfRequests = 500;
Set<Integer> sink = new HashSet<>();
FluxMessageChannel fluxMessageChannel = MessageChannels.flux().get();
fluxMessageChannel.subscribeTo(
Flux
.fromStream(IntStream
.range(0, numberOfRequests).boxed()
.map(i -> MessageBuilder.withPayload(i).build())
)
.map(Message::getPayload)
.map(sink::add)
.flatMap(i -> Mono.error(new Exception("whatever")))
);
Flux
.from(fluxMessageChannel)
.subscribe();
Thread.sleep(3000);
Assert.assertEquals(numberOfRequests, sink.size());
}
EDIT:
I just raised an issue in the reactor core project: https://github.com/reactor/reactor-core/issues/2011

Chain CompletableFuture and stop on first success

I'm consuming an API that returns CompletableFutures for querying devices (similar to digitalpetri modbus).
I need to call this API with a couple of options to query a device and figure out what it is - this is basically trial and error until it succeeds. These are embedded device protocols that I cannot change, but you can think of the process as working similar to the following:
Are you an apple?
If not, then are you a pineapple?
If not, then are you a pen?
...
While the API uses futures, in reality, the communications are serial (going over the same physical piece of wire), so they will never be executed synchronously. Once I know what it is, I want to be able to stop trying and let the caller know what it is.
I already know that I can get the result of only one of the futures with any (see below), but that may result in additional attempts that should be avoided.
Is there a pattern for chaining futures where you stop once one of them succeeds?
Similar, but is wasteful of very limited resources.
List<CompletableFuture<String>> futures = Arrays.asList(
CompletableFuture.supplyAsync(() -> "attempt 1"),
CompletableFuture.supplyAsync(() -> "attempt 2"),
CompletableFuture.supplyAsync(() -> "attempt 3"));
CompletableFuture<String>[] futuresArray = (CompletableFuture<String>[]) futures.toArray();
CompletableFuture<Object> c = CompletableFuture.anyOf(futuresArray);
Suppose that you have a method that is "pseudo-asynchronous" as you describe, i.e. it has an asynchronous API but requires some locking to perform:
private final static Object lock = new Object();
private static CompletableFuture<Boolean> pseudoAsyncCall(int input) {
return CompletableFuture.supplyAsync(() -> {
synchronized (lock) {
System.out.println("Executing for " + input);
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
return input > 3;
}
});
}
And a List<Integer> of inputs that you want to check against this method, you can check each of them in sequence with recursive composition:
public static CompletableFuture<Integer> findMatch(List<Integer> inputs) {
return findMatch(inputs, 0);
}
private static CompletableFuture<Integer> findMatch(List<Integer> inputs, int startIndex) {
if (startIndex >= inputs.size()) {
// no match found -- an exception could be thrown here if preferred
return CompletableFuture.completedFuture(null);
}
return pseudoAsyncCall(inputs.get(startIndex))
.thenCompose(result -> {
if (result) {
return CompletableFuture.completedFuture(inputs.get(startIndex));
} else {
return findMatch(inputs, startIndex + 1);
}
});
}
This would be used like this:
public static void main(String[] args) {
List<Integer> inputs = Arrays.asList(0, 1, 2, 3, 4, 5);
CompletableFuture<Integer> matching = findMatch(inputs);
System.out.println("Found match: " + matching.join());
}
Output:
Executing for 0
Executing for 1
Executing for 2
Executing for 3
Executing for 4
Found match: 4
As you can see, it is not called for input 5, while your API (findMatch()) remains asynchronous.
I think the best you can do is, after your retrieval of the result,
futures.forEach(f -> f.cancel(true));
This will not affect the one having produced the result, and tries its best to stop the others. Since IIUC you get them from an outside source, there's no guarantee it will actually interrupt their work.
However, since
this class has no direct control over the computation that causes it to be completed, cancellation is treated as just another form of exceptional completion
(from CompletableFuture doc), I doubt it will do what you actually want.

Execute different Completables in succession

I am currently trying to achieve a specific outcome with the reactive extensions in java, however I am not able to do so, maybe someone of you can help me out.
firstCompletable
.onErrorComplete(t -> specificErrorHandlingOne())
.andThen(secondCompletable())
.onErrorComplete(t -> specificErrorHandlingTwo())
.andThen(thirdCompletable())
.onErrorComplete(t -> specificErrorHandlingThree())
.andThen(fourthCompletable())
.onErrorComplete(t -> specificErrorHandlingFour())
.subscribe(viewCallback::showSuccess)
However, when there is an error in, e.g., the secondCompletable, the specific error handling is being executed but then the other Completables are still being scheduled. I want the whole chain of Completables to stop executing if one of the Completables fails. How would I do that?
I already tried using doOnError instead, but this just ended up on n a stack trace of the specific error thrown.
Completable.concat(
completable1.doOnError(e -> {...}),
completable2.doOnError(e -> {...}),
completable3.doOnError(e -> {...}),
completable4.doOnError(e -> {...})
).subscribe(action, errorConsumer);
Completables will be subscribed to in specified order
action will be invoked when all complete
You can specify error handlers for each one (this is optional)
Any error will break the pipeline and propagate to subscriber (errorConsumer)
Your original andThen chain shall also work but you need to replace onErrorComplete, which substitutes an error with completion, with doOnError, which just invokes specified action. Or just return false from your specificErrorHandlingXxx().
Try the below:
public static void main(String[] args) {
System.out.println("start");
Completable c1 = Completable.fromAction(() -> printAndWait(1, 1));
Completable c2 = Completable.fromAction(() -> printAndWait(2, 2));
Completable c3 = Completable.fromObservable(Observable.timer(3, TimeUnit.SECONDS).concatWith(Observable.error(new RuntimeException())));
Completable c4 = Completable.fromAction(() -> printAndWait(4, 2));
c1.concatWith(c2).concatWith(c3).concatWith(c4).subscribe(e -> e.printStackTrace(), () -> System.out.println("done"));
printAndWait(10, 10);//dont exit till program is completely executed
}
private static void printAndWait(int i, int j) {
System.out.println(i);
Observable.timer(j, TimeUnit.SECONDS).toBlocking().subscribe();//just add delay
}

Generate infinite sequence of Natural numbers using RxJava

I am trying to write a simple program using RxJava to generate an infinite sequence of natural numbers. So, far I have found two ways to generate sequence of numbers using Observable.timer() and Observable.interval(). I am not sure if these functions are the right way to approach this problem. I was expecting a simple function like one we have in Java 8 to generate infinite natural numbers.
IntStream.iterate(1, value -> value +1).forEach(System.out::println);
I tried using IntStream with Observable but that does not work correctly. It sends infinite stream of numbers only to first subscriber. How can I correctly generate infinite natural number sequence?
import rx.Observable;
import rx.functions.Action1;
import java.util.stream.IntStream;
public class NaturalNumbers {
public static void main(String[] args) {
Observable<Integer> naturalNumbers = Observable.<Integer>create(subscriber -> {
IntStream stream = IntStream.iterate(1, val -> val + 1);
stream.forEach(naturalNumber -> subscriber.onNext(naturalNumber));
});
Action1<Integer> first = naturalNumber -> System.out.println("First got " + naturalNumber);
Action1<Integer> second = naturalNumber -> System.out.println("Second got " + naturalNumber);
Action1<Integer> third = naturalNumber -> System.out.println("Third got " + naturalNumber);
naturalNumbers.subscribe(first);
naturalNumbers.subscribe(second);
naturalNumbers.subscribe(third);
}
}
The problem is that the on naturalNumbers.subscribe(first);, the OnSubscribe you implemented is being called and you are doing a forEach over an infinite stream, hence why your program never terminates.
One way you could deal with it is to asynchronously subscribe them on a different thread. To easily see the results I had to introduce a sleep into the Stream processing:
Observable<Integer> naturalNumbers = Observable.<Integer>create(subscriber -> {
IntStream stream = IntStream.iterate(1, i -> i + 1);
stream.peek(i -> {
try {
// Added to visibly see printing
Thread.sleep(50);
} catch (InterruptedException e) {
}
}).forEach(subscriber::onNext);
});
final Subscription subscribe1 = naturalNumbers
.subscribeOn(Schedulers.newThread())
.subscribe(first);
final Subscription subscribe2 = naturalNumbers
.subscribeOn(Schedulers.newThread())
.subscribe(second);
final Subscription subscribe3 = naturalNumbers
.subscribeOn(Schedulers.newThread())
.subscribe(third);
Thread.sleep(1000);
System.out.println("Unsubscribing");
subscribe1.unsubscribe();
subscribe2.unsubscribe();
subscribe3.unsubscribe();
Thread.sleep(1000);
System.out.println("Stopping");
Observable.Generate is exactly the operator to solve this class of problem reactively. I also assume this is a pedagogical example, since using an iterable for this is probably better anyway.
Your code produces the whole stream on the subscriber's thread. Since it is an infinite stream the subscribe call will never complete. Aside from that obvious problem, unsubscribing is also going to be problematic since you aren't checking for it in your loop.
You want to use a scheduler to solve this problem - certainly do not use subscribeOn since that would burden all observers. Schedule the delivery of each number to onNext - and as a last step in each scheduled action, schedule the next one.
Essentially this is what Observable.generate gives you - each iteration is scheduled on the provided scheduler (which defaults to one that introduces concurrency if you don't specify it). Scheduler operations can be cancelled and avoid thread starvation.
Rx.NET solves it like this (actually there is an async/await model that's better, but not available in Java afaik):
static IObservable<int> Range(int start, int count, IScheduler scheduler)
{
return Observable.Create<int>(observer =>
{
return scheduler.Schedule(0, (i, self) =>
{
if (i < count)
{
Console.WriteLine("Iteration {0}", i);
observer.OnNext(start + i);
self(i + 1);
}
else
{
observer.OnCompleted();
}
});
});
}
Two things to note here:
The call to Schedule returns a subscription handle that is passed back to the observer
The Schedule is recursive - the self parameter is a reference to the scheduler used to call the next iteration. This allows for unsubscription to cancel the operation.
Not sure how this looks in RxJava, but the idea should be the same. Again, Observable.generate will probably be simpler for you as it was designed to take care of this scenario.
When creating infinite sequencies care should be taken to:
subscribe and observe on different threads; otherwise you will only serve single subscriber
stop generating values as soon as subscription terminates; otherwise runaway loops will eat your CPU
The first issue is solved by using subscribeOn(), observeOn() and various schedulers.
The second issue is best solved by using library provided methods Observable.generate() or Observable.fromIterable(). They do proper checking.
Check this:
Observable<Integer> naturalNumbers =
Observable.<Integer, Integer>generate(() -> 1, (s, g) -> {
logger.info("generating {}", s);
g.onNext(s);
return s + 1;
}).subscribeOn(Schedulers.newThread());
Disposable sub1 = naturalNumbers
.subscribe(v -> logger.info("1 got {}", v));
Disposable sub2 = naturalNumbers
.subscribe(v -> logger.info("2 got {}", v));
Disposable sub3 = naturalNumbers
.subscribe(v -> logger.info("3 got {}", v));
Thread.sleep(100);
logger.info("unsubscribing...");
sub1.dispose();
sub2.dispose();
sub3.dispose();
Thread.sleep(1000);
logger.info("done");

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