Scala Stream processing

Question

I wrote the following code in scala using streams

def foo(x: Int) : Stream[Int] = {println("came inside"); (2 * x)} #:: foo(x + 1)
foo(1).takeWhile(_ < 6).toList.foreach(x => println(s"+++ $x"))

This works and produces the following output

came inside
came inside
came inside
+++ 2
+++ 4

but I wanted the processing to happen like

came inside
+++ 2
came inside
+++ 4
came inside

Basically I want to process one by one, till the termination condition of < 6 is met. I guess its my "tolist" method which first creates a giant list and only then processes it.

i) You're defining a stream with side-effects (i.e. println), which is hard to deal with and difficult to reason about. ii) Yes, toList will "materialize the stream", and so it will trigger all side-effects, before executing whatever you call afterwards. — rsenna
– rsenna, Commented Aug 24, 2016 at 23:02

Nathaniel Ford · Accepted Answer · 2016-08-25 00:01:02Z

1

First, format your code in a more readable fashion. Then, remove the toList: all it does is pull your entire stream into a single variable. Doing this forces all the values to be calculated. Since every value is calculated at this point, we know that the 'inside' printlns will execute before the 'outside' ones do. You want to store the definition of your function (including it's starting value) as a lazily evaluated value. This should work:

def river(x: Int) : Stream[Int] = {
  println("Inside function.")
  (2 * x) #:: river(x + 1)
}

lazy val creek = river(1)
creek.takeWhile(_ < 6).foreach(x => println(s"+++ $x"))

edited Aug 25, 2016 at 0:01

answered Aug 24, 2016 at 23:03

Nathaniel Ford

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