Question

我正在通过＆＃34; Scala中的功能编程＆＃34;书。第5.2章从以下代码开始：

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

object Stream {
  def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = {
    lazy val head = hd
    lazy val tail = tl
    Cons(() => head, () => tail)
  }

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

  def apply[A](as: A*): Stream[A] =
    if (as.isEmpty) empty else cons(as.head, apply(as.tail: _*))

}

然后给出以下定义：

  def headOption: Option[A] = this match {
      case Empty => None
      case Cons(h, t) => Some(h())
    }

我认为它应该是伴侣对象的一部分。为了不立即得到编译错误，我添加＆＃34; headOption [A]＆＃34;。但是当我将结果包含在＆＃34;对象流＆＃34;我收到错误＆＃34;模式类型与预期类型不兼容; found：myPackage.Empty.type required：myPackage.Stream.type＆＃34;与＆＃34;空＆＃34;下划线。

我在这里有点失落。我做错了什么？

Answer 1

它应该是trait Stream[+A]。

的一部分

此处this引用Stream（特征流不是伴侣对象）。 Stream可以是Empty，也可以是Cons。因此，this上的模式匹配是有意义的。

headOption给出了流的第一个元素（如果可用）。

    case object Empty extends Stream[Nothing]

    case class Cons[+A](h: () => A, t: () => Stream[A]) extends Stream[A]

    sealed trait Stream[+A] {
      def headOption: Option[A] = this match {
        case Empty => None
        case Cons(h, t) => Some(h())
      }
    }



    object Stream {

      def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = {
        lazy val head = hd
        lazy val tail = tl
        Cons(() => head, () => tail)
      }

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

      def apply[A](as: A*): Stream[A] =
        if (as.isEmpty) empty else cons(as.head, apply(as.tail: _*))


    }

Scala REPL

scala> :paste
// Entering paste mode (ctrl-D to finish)

case object Empty extends Stream[Nothing]

case class Cons[+A](h: () => A, t: () => Stream[A]) extends Stream[A]

sealed trait Stream[+A] {

  def headOption: Option[A] = this match {
    case Empty => None
    case Cons(h, t) => Some(h())
  }

}

object Stream {

  def cons[A](hd: => A, tl: => Stream[A]): Stream[A] = {
    lazy val head = hd
    lazy val tail = tl
    Cons(() => head, () => tail)
  }

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

  def apply[A](as: A*): Stream[A] =
    if (as.isEmpty) empty else cons(as.head, apply(as.tail: _*))

}

// Exiting paste mode, now interpreting.

defined object Empty
defined class Cons
defined trait Stream
defined object Stream

scala> Stream.cons(1, Empty).headOption
res0: Option[Int] = Some(1)

scala错误＆＃34;模式类型与预期类型不兼容＆＃34;

1 个答案: