Lazy Evaluation in a Stream?

4
votes

Given a lazy val:

scala> lazy val y = {println("Y!"); 200}
y: Int = <lazy>

I tried to put y into a Stream - to find out if it would be eagerly or lazily evaluated.

scala> Stream(100, y)
Y!
res4: scala.collection.immutable.Stream[Int] = Stream(100, ?)

Clearly it's eagerly evaluated.

Besides the following, how can I create a Stream that lazily evaluates its members?

scala> Stream[() => Int](() => 100, () => 200)
res18: scala.collection.immutable.Stream[() => Int] = Stream(<function0>, ?)

scala> res18.map(_())
res19: scala.collection.immutable.Stream[Int] = Stream(100, ?)

scala> res19.last
res20: Int = 200

scala> res19
res21: scala.collection.immutable.Stream[Int] = Stream(100, 200)
2
scalastream

2 Answers

5
votes

Stream.apply takes a varargs parameter, and it's not possible to have by-name varargs parameters in Scala. You can use the #:: syntax for streams, though:

scala> lazy val y = {println("Y!"); 200}
y: Int = <lazy>

scala> val s = 100 #:: y #:: Stream.empty
s: scala.collection.immutable.Stream[Int] = Stream(100, ?)

scala> s.last
Y!
res0: Int = 200

This works because the ConsWrapper class and implicit conversion that are used to add #:: to streams both take a by-name parameter.

0
votes

Travis is right, but just for kicks:

scala> implicit def a2f[T](t : => T) : (() => T) = () => t
a2f: [T](t: => T)() => T

scala> def lazyStream[T]( args : (() => T)* ) : Stream[T] = Stream(args:_*).map(_())
lazyStream: [T](args: () => T*)Stream[T]

scala> lazy val y = {println("Y!"); 200}
y: Int = <lazy>

scala> lazyStream(100, y)
res8: Stream[Int] = Stream(100, ?)

scala> res8.last
Y!
res9: Int = 200