Riddle number 6 does not check for parallelism about rxriddles HOT 5 CLOSED

I don't quite understand the problem.

RxRule is having a Scheduler that will run. The essential part is that subscribe is called eagerly. Which is asserted with assertEmpty() & assertThat(subscribeCounter.get()).isEqualTo(2) // We want to subscribe immediately.

from rxriddles.

I'm going to close this for now as I don't see a reason to change this. If you feel otherwise feel free to message me again.

from rxriddles.

@vanniktech I'll message here since I had similar thoughts when doing riddle no 6.

The point is, the solution proposed:

object Riddle6Solution {
  fun solve(first: Single<Int>, second: Single<Int>)
      = Single.zip(first, second, BiFunction<Int, Int, Pair<Int, Int>> { t1, t2 -> t1 to t2 })
}

does not introduce any parallelism. The comment states:

Execute both [first] and [second] Single's in parallel and provide both results as a pair.

but the solution does not execute any arbitrary pair of singles in parallel. On the contrary, as long as first and single doesn't specify schedulers, the solution works sequentially. The solution is appropriate only for this specific use case (implemented in tests), where singles already work on separate scheduler(s).

For actual educational value I'd expect that the provided singles don't have schedulers applied. I believe this is one of the things that are surprising for the beginners: in order to have the two streams work in parallel, we need to move them off the main thread manually, while the riddle suggests otherwise

from rxriddles.

Yup, that's right! Should we add a sentence to the documentation that we can assume that each Single has already a Scheduler applied? Or how do you want to proceed?

from rxriddles.

@vanniktech took me a while to find time for this, here is my suggestion. I believe not providing a scheduler and not changing the comment is the way to go for this to still be a riddle ;) Let me know what you think

from rxriddles.