使用 delayElements 时程序在 Flux 完成之前终止

Question

I'm learning about reactive programming but I have a doubt, I'm trying to execute the following code but I don't understand why if I add delayElements before subscribe the elements are not printed

Without delayElements the values are shown

import reactor.core.publisher.Flux;

import java.time.Duration;

public class PlayWithFlux {

    public static void main(String[] args){
        Flux<Integer> flux = Flux.just(1, 2, 3);
        flux.log().subscribe(System.out::println);
    }
}

With delayElements the elements are not shown

import reactor.core.publisher.Flux;

import java.time.Duration;

public class PlayWithFlux {

    public static void main(String[] args){
        Flux<Integer> flux = Flux.just(1, 2, 3).delayElements(Duration.ofSeconds(1));
        flux.log().subscribe(System.out::println);
    }
}

Is there an explanation about this behavior?

Answer 1

The problem here is that delayElements does as it says, it delays before emitting elements.

Calling subscribe is not a blocking event, its an async, or "fire and forget" operator.

So what is actually happening is that you call subscribe then the main flow continues and ends the program, before any values have time to get emitted.

To solve this just quick and dirty is to delay the end of the program with for instance a call to Thread.sleep()

public static void main(String[] args){
    Flux.just(1, 2, 3)
        .delayElements(Duration.ofSeconds(1))
        .subscribe(System.out::println);

    // Delay program shutdown so we can watch the elements get printed
    Thread.sleep(5000);
}

Answer 2

If you are only using Flux.just and subscribe , then everything is executed on the main thread, so the program will terminate only after the Flux is finished.

Once delayElements steps into the picture, the Flux is no longer executed on the main thread. Internally delayElements switches to a so-called parallel scheduler, which is basically a thread pool. Since now the Flux is running on a different thread, the main thread will continue the execution without waiting for the Flux to finish and the program will terminate before the Flux would finish.

Instead of using Thread.sleep , it is better to use one of the block... methods in this case. This way the program can finish right after the Flux finished:

public static void main(String[] args){
    Flux.just(1, 2, 3)
        .delayElements(Duration.ofSeconds(1))
        .doOnNext(System.out::println)
        .blockLast();
}