流中的懒惰评估？

Question

鉴于lazy val:

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

我试图y进入Stream- 以确定是否会急切或懒惰地评估.

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

显然,它受到了热切的评价.

除了以下内容,我如何创建一个Stream懒惰的评估其成员？

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.apply采用varargs参数,并且在Scala中不可能有名称的varargs参数.但是,您可以使用#::流的语法:

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

这是有效的,因为ConsWrapper用于添加#::到流的类和隐式转换都采用名称参数.