Add Source.interleaveAll combinator #13
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rucek
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| * scala> val res0: List[Int] = List(1, 2, 10, 20, 100, 200, 3, 4, 30) | ||
| * }}} | ||
| */ | ||
| def interleaveAll[U >: T](others: Seq[Source[U]], segmentSize: Int = 1, eagerComplete: Boolean = false)(using |
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since none of the sources is special, maybe this shoul dbe a method on Source - Source.interlaveAll?
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One downside is that this would require us to handle the case of an empty list - but then I think we just return a done source.
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Good point with moving it to Source. As for the empty list - we can either return a done source as you suggest, or change the signature of interleaveAll so that it enforces at least two sources (since one source would be a special case as well, where we just return the argument), e.g.
def interleaveAll[T](first: Source[T], second: Source[T], rest: Source[T]*)(segmentSize: Int, eagerComplete: Boolean)The downside of the latter approach would be making the API inconsistent between interleave and interleaveAll. WDYT?
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I don't think there's an inconsistency, interleave acts on two sources using the more convenient dot-notation, the ...All variant takes an arbitrary list. The first / second approach would be good, but then it enforces static structure - won't work if you have a dynamically-defined list of sources.
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