Is it possible to use 'yield' to generate 'Iterator' instead of a list in Scala?

Question

Is it possible to use yield as an iterator without evaluation of every value?

It is a common task when it is easy to implement complex list generation, and then you need

Answer 1

I have a List...

scala>  List(1, 2, 3)
res0: List[Int] = List(1, 2, 3)

And a function...

scala> def foo(i : Int) : String = { println("Eval: " + i); i.toString + "Foo" }
foo: (i: Int)String

And now I'll use a for-comprehension with an Iterator...

scala> for { i <- res0.iterator } yield foo(i)
res2: Iterator[java.lang.String] = non-empty iterator

You can use a for comprehension on any type with flatMap, map and filter methods. You could also use the views:

scala> for { i <- res0.view } yield foo(i)
res3: scala.collection.SeqView[String,Seq[_]] = SeqViewM(...)

Evaluation is non-strict in either case...

scala> res3.head
Eval: 1
res4: String = 1Foo