How to chain Iterable without computing underlying iterators

Question

If I had two iterators I could just write iter1 ++ iter2 and iterators would not be computed until they are needed. Is there any way to chain Iterable instances in the same way?

I tried to use iterable1 ++ iterable2 but it causes immediately calculating nested values just like they are added to some structure. Is it possible to avoid this extra calculations and creating extra data structures?

I actually think that's kind of the purpose of iterators in the first place :D So I'd use iterable1.iterator ++ iterable2.iterator to avoid computation. — C4stor
– C4stor, Commented Mar 7, 2017 at 14:16
If you already have Iterable, than you've already calculated your nested values. It's not the chaining that is causing the caculations. — Jon Anderson
– Jon Anderson, Commented Mar 7, 2017 at 14:38

Jasper-M · Accepted Answer · 2017-03-07 14:42:05Z

1

No. Iterable is just an interface that can be implemented by anything that can be iterated over. So when you have an Iterable[Int] that can be either a lazy or a strict collection, there's no way to know.

scala> val iterable1: Iterable[Int] = List(1,2,3)
iterable1: Iterable[Int] = List(1, 2, 3)

scala> iterable1 ++ iterable1
res2: Iterable[Int] = List(1, 2, 3, 1, 2, 3)

scala> val iterable2: Iterable[Int] = List(1,2,3).view
iterable2: Iterable[Int] = SeqView(...)

scala> iterable2 ++ iterable2
res3: Iterable[Int] = SeqViewA(...)

answered Mar 7, 2017 at 14:42

Jasper-M

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2 Comments

ayvango Over a year ago

So, I should basically use iterable.view to get desired behaviour?

Jasper-M Over a year ago

@ayvango Yes, I think so. Or iterable.iterator. Documentation says that "all views for iterable collections are defined by re-interpreting the iterator method".

Juh_ · Accepted Answer · 2018-03-12 12:59:44Z

0

You can code a simple chain-iterable that is a lazy concatenation of Iterable:

case class Chain[A](iter: Iterable[A]*) extends Iterable[A]{
  def iterator: Iterator[A] = iter.map(_.iterator).foldLeft(Iterator.empty: Iterator[A])(_++_)
}

Chain(List(1,2,3), 10 to 15, Vector(42,13))

answered Mar 12, 2018 at 12:59

Juh_

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