Scala中的"call-cc"模式？

Question

我发现了一篇关于call with current continuation模式的好文章.据我所知,他们使用Scheme和undelimited continuations.文章中的模式可以在Scala中实现吗？是否有关于delimited continuationsScala模式的文章？

Answer 1

是的,他们绝对可以.callCC在Scala中看起来像这样:

def callCC[R, A, B](f: (A => Cont[R, B]) => Cont[R, A]): Cont[R, A] =
  Cont(k => f(a => Cont(_ => k(a))) run k)

Cont捕获延续的数据结构在哪里:

case class Cont[R, A](run: (A => R) => R) {
  def flatMap[B](f: A => Cont[R, B]): Cont[R, B] =
    Cont(k => run(a => f(a) run k))
  def map[B](f: A => B): Cont[R, B] =
    Cont(k => run(a => k(f(a))))
}

以下是如何使用它来模拟已检查的异常:

def divExcpt[R](x: Int, y: Int, h: String => Cont[R, Int]): Cont[R, Int] =
  callCC[R, Int, String](ok => for {
    err <- callCC[R, String, Unit](notOK => for {
             _ <- if (y == 0) notOK("Denominator 0") else Cont[R, Unit](_(()))
             r <- ok(x / y)
           } yield r)
    r <- h(err)
  } yield r)

您可以按如下方式调用此函数:

scala> divExcpt(10, 2, error) run println   
5

scala> divExcpt(10, 0, error) run println
java.lang.RuntimeException: Denominator 0