scala 前向引用扩展了定义

Question

case object Empty extends Stream[Nothing]
case class Cons[+A](h: () => A, t: () => Stream[A]) extends Stream[A]

sealed trait Stream[+A] {

  def toList: List[A] = {
    val buf = new collection.mutable.ListBuffer[A]
    def go(s: Stream[A]): List[A] = s match {
      case Cons(h, t) =>
        buf += h()
        go(t())
      case _ => buf.toList
    }
    go(this)
  }

  def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = Stream.cons(hd, tl)

  def empty = Stream.empty

  def unfold[A, S](z: S)(f: S => Option[(A, S)]): Stream[A] = f(z) match {
    case Some((h,t)) => cons(h, unfold(t)(f))
    case None => empty
  }

  def take(n: Int): Stream[A] = unfold((this, n)) {
    case (Cons(h, t), 1) => Some((h(), (empty, 0)))
    case (Cons(h, t), n) if n > 1 => Some((h(), (t(), n-1)))
    case (Empty, 0) => None
  }
}

object Stream {

  def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = Cons(() => hd, () => tl)

  def empty[A]: Stream[A] = Empty

  val ones: Stream[Int] = Stream.cons(1, ones)
}

object StreamTest {
  def main(args: Array[String]) {

    //why compile error: forward reference extends over definition of value ones
    /*val ones: Stream[Int] = Stream.cons(1, ones)
    println(ones.take(5).toList)*/

    println(Stream.ones.take(5).toList)
  }
}

为什么编译错误？：前向引用扩展了值的定义

在对对象“流”中， val 个：Stream[Int] = Stream.scons(1, one) 没问题

但在 main 方法中，这是不行的（但是...相同的合成物！）

Answer 1

前向引用在Cons[+A]...这行引用：

def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = Cons(() => hd, () => tl)

尝试移动

case object Empty extends Stream[Nothing]
case class Cons[+A](h: () => A, t: () => Stream[A]) extends Stream[A]

进入伴随对象。

Answer 2

本地 val 不是成员。

对于您的测试代码，请执行以下操作：

object StreamTest extends App {
  //def main(args: Array[String]) {

    //why compile error: forward reference extends over definition of value ones
    val ones: Stream[Int] = Stream.cons(1, ones)
    println(ones.take(5).toList)

    println(Stream.ones.take(5).toList)
  //}
}

block 中的限制用 in the spec here 措辞，其中另一个建议是让它成为 block 中的惰性 val，具有相同的效果。