Flux.zip method not emitting all elements

Question

I am working with Reactive Stream and Publishers (Mono and Flux), and combining the two publishers using the zip and zipWith method of Flux as follows:

Flux<String> flux1 = Flux.just(" {1} ","{2} ","{3} ","{4} " );
Flux<String> flux2 = Flux.just(" |A|"," |B| "," |C| ");
Flux.zip(flux1, flux2,
                    (itemflux1, itemflux2) -> "[ "+itemflux1 + ":"+ itemflux2 + " ] " )
            .subscribe(System.out::print);

and here is the output:

[  {1} : |A| ] [ {2} : |B|  ] [ {3} : |C|  ]

AS flux1 has four elements and flux2 has three elements, the forth element in flux1 gets lost. And when i tried to print the logs of the flux, there is no information about what happened to the forth element.

Here is the statement for printing logs:

Flux.zip(flux1, flux2,
                (itemflux1, itemflux2) -> "[ "+itemflux1 + ":"+ itemflux2 + " ] " ).log()
        .subscribe(System.out::print);

and here is the console logs with using log method:

[info] onSubscribe(FluxZip.ZipCoordinator)
[info] request(unbounded)
[info] onNext([  {1} : |A| ] )
[  {1} : |A| ] [info] onNext([ {2} : |B|  ] )
[ {2} : |B|  ] [info] onNext([ {3} : |C|  ] )
[ {3} : |C|  ] [info] onComplete()

From the documentation of zip method, i got

The operator will continue doing so until any of the sources completes. Errors will immediately be forwarded. This "Step-Merge" processing is especially useful in Scatter-Gather scenarios.

But in my case, it did not logged any error and did not logged any message about the lost element.

How can i get the information about the lost element?

Please suggest.

score 4 · Accepted Answer · answered Jan 03 '18 at 11:10

zip/zipWith will output as many pairs as there are elements in the shortest Flux. It cancels the longer Flux upon termination of the smallest one, which should be visible if you put the log() on the source Flux instead of the zipped one.

This is demonstrated by this snippet (which is tuned to show 1-by-1 requests and run as a unit test, hence the hide()/zipWith(..., 1) and blockLast()):

@Test
public void test() {
    Flux<Integer> flux1 = Flux.range(1, 4).hide().log("\tFLUX 1");
    Flux<Integer> flux2 = Flux.range(10, 2).hide().log("\tFlux 2");

    flux1.zipWith(flux2, 1)
        .log("zipped")
        .blockLast();
}

Which outputs:

11:57:21.072 [main] INFO  zipped - onSubscribe(FluxZip.ZipCoordinator)
11:57:21.077 [main] INFO  zipped - request(unbounded)
11:57:21.079 [main] INFO    FLUX 1 - onSubscribe(FluxHide.HideSubscriber)
11:57:21.079 [main] INFO    FLUX 1 - request(1)
11:57:21.079 [main] INFO    FLUX 1 - onNext(1)
11:57:21.079 [main] INFO    Flux 2 - onSubscribe(FluxHide.HideSubscriber)
11:57:21.080 [main] INFO    Flux 2 - request(1)
11:57:21.080 [main] INFO    Flux 2 - onNext(10)
11:57:21.080 [main] INFO  zipped - onNext([1,10])
11:57:21.080 [main] INFO    FLUX 1 - request(1)
11:57:21.080 [main] INFO    FLUX 1 - onNext(2)
11:57:21.080 [main] INFO    Flux 2 - request(1)
11:57:21.080 [main] INFO    Flux 2 - onNext(11)
11:57:21.080 [main] INFO  zipped - onNext([2,11])
11:57:21.080 [main] INFO    FLUX 1 - request(1)
11:57:21.080 [main] INFO    FLUX 1 - onNext(3)
11:57:21.080 [main] INFO    Flux 2 - request(1)
11:57:21.080 [main] INFO    Flux 2 - onComplete()
11:57:21.081 [main] INFO    FLUX 1 - cancel() <----- HERE
11:57:21.081 [main] INFO    Flux 2 - cancel()
11:57:21.081 [main] INFO  zipped - onComplete()

This is the "until any of the sources completes" part.

Could you please suggest the method, which will continue zipping remaining publishers? Or how we can get around with this situation? — incompletablefuture, Apr 05 '21 at 16:31

score 2 · Answer 2 · answered Jan 02 '18 at 10:52

This is the expected behavior of this operator. With Flux.zip, one of the provided Flux might be an infinite one; a common example of that is zipping a Flux of data with a Flux.interval(Duration duration) instance (which is infinite).

If you're stuck in this situation, it probably means that you need to use a different operator.

score 0 · Answer 3 · answered Jan 02 '18 at 11:01

To explain what the documentation says
Errors will immediately be forwarded. --> Means if there is an error in the combinator function then it is immediately forwarded. You can check this by making one of the entries in either of the Flux as null.

There is no way to get the lost element. Because the stream is not read further when one of the Stream has ended. Hope it is clear. If you really want to get the last element of the stream, try other operators.

Flux.zip method not emitting all elements

3 Answers3