Question

我根据自身定义了一个流（递归定义）。尝试访问流的第二个元素时，会抛出StackOverflowError。来自scala控制台的代码：

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$$anonfun$apply$1.apply(<console>:9)
        at $anonfun$s2$1$$anonfun$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...

我无法理解堆栈溢出的原因。由于流本质上是惰性的，因此递归映射应该有效。

这种情况有什么问题？

我使用的是Scala版本2.8.0.RC2。

Answer 1

问题是zipped不是懒惰的 - 它实际上试图在那里评估该地图。您可以使用zip执行所需操作。

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

递归流抛出StackOverflowError

1 个答案: