Lazy appreciation in the stream?

Question

Given lazy val:

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

I tried putting yin Stream- to find out if he would be impatiently or lazily appreciated.

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

Clearly, he looked forward to.

Also, how can I create Streamone that lazily values ​​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)

Answer 1

Stream.applyaccepts the varargs parameter and cannot have the varargs parameters by name in Scala. You can use the syntax #::for threads:

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 class ConsWrapperand implicit conversion that are used to add #::to the streams accept the by-name parameter.

Answer 2

, :

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