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So I have a code that looks something like this:

let a = Observable.just("A").delay(.seconds(3), scheduler: MainScheduler.instance)
let b = Observable.just("B").delay(.seconds(3), scheduler: MainScheduler.instance)
Observable.merge([a, b]).subscribe(onNext: { print($0) })

I thought that printed order should be random, as the delay that finishes marginally earlier will be emitted first by merge. But the order seems to be determined solely by the order of variables in array passed to merge.

That means .merge([a, b]) prints

A
B

but .merge([b, a]) prints

B
A

Why is this the case? Does it mean that merge somehow handles the concurrent events and I can rely on the fact that the events with same delay will always be emitted in their order in array?

I know I could easily solve it by using .from(["A", "B"]), but I am now curious to know how exactly does the merge work.

MrKew
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1 Answers1

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You put both delays on the same thread (and I suspect the subscriptions are also happening on that thread,) so they will execute on the same thread in the order they are subscribed to. Since the current implementation of merge subscribes to the observables in the order it receives them, you get the behavior you are seeing. However, it's important to note that you cannot rely on this, there is no guarantee that it will be true for every version of the library.

Daniel T.
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