Flux的方法publishOn没有按预期工作

Question

我正试图将一个阻塞消费者集成为Reactor铝-SR1中的助焊剂订户。我想使用一个并行调度器，并发地执行阻塞操作。

我实现了一个主类来描述我意图：

package etienne.peiniau;

import org.reactivestreams.Subscriber;
import org.reactivestreams.Subscription;
import reactor.core.publisher.Flux;
import reactor.core.scheduler.Schedulers;
import reactor.util.function.Tuple2;

public class Main {

    public static void main(String[] args) throws InterruptedException {
        Flux.just(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20)
                .elapsed()
                .publishOn(Schedulers.parallel())
                .subscribe(new Subscriber<Tuple2<Long, Integer>>() {
                    @Override
                    public void onSubscribe(Subscription subscription) {
                        System.out.println("[" + Thread.currentThread().getName() + "] Subscription");
                        subscription.request(Long.MAX_VALUE);
                    }

                    @Override
                    public void onNext(Tuple2<Long, Integer> t2) {
                        System.out.println("[" + Thread.currentThread().getName() + "] " + t2);
                        try {
                            Thread.sleep(1000); // long operation
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }

                    @Override
                    public void onError(Throwable throwable) {
                        System.err.println("[" + Thread.currentThread().getName() + "] Error: " + throwable.getMessage());
                    }

                    @Override
                    public void onComplete() {
                        System.out.println("[" + Thread.currentThread().getName() + "] Complete");
                    }
                });
        // Waiting for the program to complete
        System.out.println("[" + Thread.currentThread().getName() + "] Main");
        Thread.sleep(100000);
    }

}

[main] Subscription
[main] Main
[parallel-1] [3,1]
[parallel-1] [1000,2]
[parallel-1] [1001,3]
[parallel-1] [1000,4]
[parallel-1] [1000,5]
[parallel-1] [1000,6]
[parallel-1] [1001,7]
[parallel-1] [1000,8]
[parallel-1] [1000,9]
[parallel-1] [1000,10]
[parallel-1] [1000,11]
[parallel-1] [1001,12]
[parallel-1] [1000,13]
[parallel-1] [1000,14]
[parallel-1] [1000,15]
[parallel-1] [1000,16]
[parallel-1] [1001,17]
[parallel-1] [1000,18]
[parallel-1] [1000,19]
[parallel-1] [1000,20]
[parallel-1] Complete

Answer 1

实际上，它如预期的那样工作。您可以看到，在并行运行时间内处理的所有值几乎相同，但您总是在相同的线程中接收元素，并且每次等待1秒时都是这样。

我想在简单的fluxparallel并不意味着更多的线程，它意味着并行工作。例如，如果您运行如下代码：

Flux.fromIterable(IntStream.range(0, 20).boxed().collect(Collectors.toList()))
            .map(i -> {
                System.out.println("map [" + Thread.currentThread().getName() + "] " + i);
                return i;
            })
            .elapsed()
            .publishOn(Schedulers.single())
            .subscribeOn(Schedulers.single())
            .subscribe(t2 -> {
                System.out.println("subscribe [" + Thread.currentThread().getName() + "] " + t2);
            });

您将看到结果：

map [single-1] 0
map [single-1] 1
...
subscribe [single-1] [4,0]
subscribe [single-1] [0,1]
...

map [single-1] 3
subscribe [parallel-1] [5,0]
map [single-1] 4
subscribe [parallel-1] [0,1]
map [single-1] 5
...

Flux.fromIterable(IntStream.range(0, 20).boxed().collect(Collectors.toList()))
        .elapsed()
        .parallel()
        .runOn(Schedulers.parallel())
        .subscribe(t2 -> {
            System.out.println("[" + Thread.currentThread().getName() + "] " + t2);
            try {
                Thread.sleep(1000); // long operation
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }, throwable -> {
            System.err.println("[" + Thread.currentThread().getName() + "] Error: " + throwable.getMessage());
        }, () -> {
            System.out.println("[" + Thread.currentThread().getName() + "] Complete");
        }, subscription -> {
            System.out.println("[" + Thread.currentThread().getName() + "] Subscription");
            subscription.request(Long.MAX_VALUE);
        });

[parallel-1] [8,0]
[parallel-2] [0,1]
[parallel-3] [0,2]
[parallel-4] [0,3]
[parallel-1] [0,4]
...

因此它只使用很少的线程来处理结果。在我看来，这确实是平行的。

还要注意，如果使用方法.subscribe(subscriber s)，所有结果都将以顺序方式使用，若要并行使用它们，应使用：

public void subscribe(Consumer<? super T> onNext, Consumer<? super Throwable>
            onError, Runnable onComplete, Consumer<? super Subscription> onSubscribe)

或任何其他具有使用者<的重载方法？超级t>onNext，...参数