RxJava链式调用中多subscribeOn()的作用及影响咨询

Great question! Let's break down how subscribeOn() works in RxJava and clarify your scenario step by step.

Core Rules of subscribeOn() in RxJava #

First, let's get the key behavior straight:

• subscribeOn() defines the thread used to subscribe to the upstream observable chain. This subscription signal travels from your subscribe() call all the way up to the original source (like your fromCallable), so the thread you pick affects how all upstream operations run.
• When you have multiple subscribeOn() calls in the same chain, the one closest to your subscribe() call (the last one in the chain) takes precedence. Earlier subscribeOn() calls are effectively ignored because the subscription thread gets switched again as the signal moves downstream to upstream.

Analyzing Your Code #

Let's look at your exact chain:

Single.fromCallable { repository.apiCall1() }
    .subscribeOn(Schedulers.io()) // This is overridden
    .flatMap { result -> 
        Single.fromCallable { repository.apiCall2() } 
    }
    .subscribeOn(Schedulers.io()) // This is the effective one
    .map { /* do something */ }
    .observeOn(Schedulers.ui())

1. Do the two API calls run on different threads?

No, they don't. Here's the breakdown:

• The second subscribeOn(Schedulers.io()) is the one that sets the thread context for the entire main chain. Both apiCall1() (from the source Single) and apiCall2() (inside flatMap) will run on threads from the Schedulers.io() pool.
• The inner Single in flatMap doesn't have its own subscribeOn(), so it inherits the thread from the main chain's subscription context.
• A quick note: They might run on different threads within the io pool (since Schedulers.io() uses a cached thread pool), but they're still part of the same scheduler—so not "different threads" in the sense of distinct background/UI threads.

If you wanted apiCall2() to run on a truly separate thread (e.g., a new thread instead of the io pool), you'd add a subscribeOn() directly to the inner Single:

.flatMap { result -> 
    Single.fromCallable { repository.apiCall2() }
        .subscribeOn(Schedulers.newThread()) // Now apiCall2 runs on a dedicated new thread
}

2. What happens if we remove the first subscribeOn(Schedulers.io())?

Absolutely nothing changes in terms of thread behavior. The second subscribeOn(Schedulers.io()) is still the effective one, so both API calls will still run on Schedulers.io() threads. The first subscribeOn() was already being overridden by the second one, so removing it has no impact.

Just to clarify: If you removed both subscribeOn() calls, that would be a problem—your API calls would run on the thread that calls subscribe(). If you subscribe on the UI thread, this would block the UI (a big no-no for Android apps). That's why we use subscribeOn() to offload heavy work like network calls to background threads.

内容的提问来源于stack exchange，提问作者Bootstrapper