Question

Why is it that,

scala> List(1,2,3,4).iterator.map((x: Int) => println(x))

does not print out

1
2
3
4

while

List(1,2,3,4).map((x: Int) => println(x))
List(1,2,3,4).foreach((x: Int) => println(x))
List(1,2,3,4).iterator.foreach((x: Int) => println(x))

all do?

In other words, why is it that a map on a iterator that maps type T to Unit and has side effects unable to show those side effects?

Edit:

Also why does the following invocation of lazyMap actually computes the new iterator (provide the complete new iterator) from beginning to end if iterator is lazy?

def lazyMap[T, U](coll: Iterable[T], f: T => U) = new Iterable[U] {
  def iterator = coll.iterator map f
}

scala> lazyMap(List(1,2,3,4), (x: Int) => x + 1)
res4: java.lang.Object with Iterable[Int] = (2, 3, 4, 5)

Answer 1

Cause map on iterator is lazy and you need some strictness:

scala> List(1,2,3,4).iterator.map((x: Int) => println(x))
res0: Iterator[Unit] = non-empty iterator

// nothing actually happened yet, just remember to do this printing things

scala> res0.toList
1
2
3
4
res1: List[Unit] = List((), (), (), ())

When you doing foreach on iterator it is quite obvious that you’re doing side effects, so lazyness will be undesired. I wouldn’t said so about map.

UPD

As for your edit: the reason for such behaviour, is that there is implicit call of toString for statement result which in turn stricts the iterator — try this code on your own:

scala> { lazyMap(List(1,2,3,4), {(x: Int) => println(x); x + 1}); 1 }

and you’ll see that function f is never called