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Recursive stream throws StackOverflowError

开发者 https://www.devze.com 2023-01-01 19:22 出处:网络
I am defining a stream in terms of itself (a recursive definition). When trying to access the second element of the stream, StackOverflowError is thrown. The code from scala console:

I am defining a stream in terms of itself (a recursive definition). When trying to access the second element of the stream, StackOverflowError is thrown. The code from scala console:

scala> val s1 = Stream.iterate(1)(identity _)
s1: scala.collection.immutable.Stream[Int] = Stream(1, ?)

scala> lazy val s2 : Stream[Int]= Stream.cons(1, (s2, s1).zipped.map { _ + _ })
s2: Stream[Int] = <lazy>

scala> s1 take 5 print
1, 1, 1, 1, 1, empty
scala> s2(0)
res4: Int = 1

scala> s2(1)
java.lang.StackOverflowError
        at $anonfun$s2$1$$anonfun$apply$1.apply(<console>:9)
        at $anonfun$s2$1$$anon开发者_Python百科fun$apply$1.apply(<console>:9)
        at scala.Tuple2$Zipped$$anonfun$map$1.apply(Tuple2.scala:62)
        at scala.collection.immutable.Stream.foreach(Stream.scala:255)
        at scala.Tuple2$Zipped.map(Tuple2.scala:60)
        at $anonfun$s2$1.apply(<console>:9)
        at $anonfun$s2$1.apply(<console>:9)
        at scala.collection.immutable.Stream$Cons.tail(Stream.scala:556)
        at scala.collection.immutable.Stream$Cons.tail(Stream.scala:550)
        at scala.collection.immutable.Stream.foreach(Stream.scala:256)
        at scala.Tuple2$Zipped.map(Tuple2.scala:60)
        at $anonfun$s2$1.apply(<console>:9)
        at $anonfun$s2$1.apply(<console>:9)
        at scala.collection.immutable.Stream$Cons.tail(Stream.scala:556)
        at scala.collection.immutable.Str...

I am unable to understand the reason of the stack overflow. Since streams are inherently lazy, recursive mapping should work.

What is wrong with this scenario?

I am using Scala version 2.8.0.RC2.


The problem is that zipped is not lazy--it actually tries to evaluate that map right there. You can do what you want with zip.

scala> val s1 = Stream.iterate(1)(identity _)
s1: scala.collection.immutable.Stream[Int] = Stream(1, ?)

scala> lazy val s2: Stream[Int] = Stream.cons(1, (s2 zip s1).map {s=>s._1+s._2})
s2: Stream[Int] = <lazy>

scala> s2 take 5 print
1, 2, 3, 4, 5, empty
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