RXJava2阅读器可用于并发订阅者

Question

I'm trying to implement an RXJava2 Observable<String> off a BufferedReader . So far so good:

public class Launcher {
    public static void main(String[] args) throws Exception {

        Process process = Runtime.getRuntime().exec("ls /tmp");
        BufferedReader reader = new BufferedReader (new InputStreamReader(process.getInputStream()));

        Observable source = Observable.create(emitter -> {
            String line = "";
            while (line != null) {
                line = reader.readLine();
                if (line == null)
                    break;
                System.out.println("Engine " + Thread.currentThread().getName() + " - " + line); //process line
                emitter.onNext(line);
            }
            emitter.onComplete();
        }).subscribeOn(Schedulers.newThread());

        source.subscribe(line -> System.out.println("UI 1   " + Thread.currentThread().getName() + " - " + line));
        source.subscribe(line -> System.out.println("UI 2   " + Thread.currentThread().getName() + " - " + line));

        TimeUnit.SECONDS.sleep(10);

    }
}

onSubscribe makes subscribers be notified in a parallel fashion. Which, if I'm not mistaken, means that the lambda in create() will be executed in parallel for each consumer.

As a consequence, if I have two subscribers, each of them gets half of the lines of the reader. Thread-1 calls readLine() that gets a line Thread-2 will not get, just the next one.

This all makes sense, still, I must be missing something, because I can't figure out how to:

1. read the lines in one thread
2. notify all subscribers concurrently - so each gets all lines

I looked into Subject s, tried to chaining Observable s, still couldn't figure it out yet.

Edit: I updated the example to a full runnable class. From what I understand the issue is hot vs cold Observables. As if the docs said Observable.create(...) should create a cold one, whereas my code clearly behaves as hot.

Follow-up question: if I add the type parameter making it Observable<String> then the onSubscribe call breaks the code, and it won't compile, as that would return Observable<Object> . Why? Calling onSubscribe on an intermediate parameter oddly works:

Observable<String> source = Observable.create(emitter -> {...});
Observable<String> source2 = source.subscribeOn(Schedulers.newThread())

Answer 1

Use publish :

ConnectableObservable<String> o = Observable.create(emitter -> {
    try (BufferedReader reader = ...) {
        while (!emitter.isDisposed()) {
            String line = reader.readLine();
            if (line == null || line.equals("end")) {
                emitter.onComplete();
                return;          
            }
            emitter.onNext(line);
        }
    }
}).subscribeOn(Schedulers.io())
  .publish();

o.subscribe(/* consumer 1 */);
o.subscribe(/* consumer 2 */);

o.connect();