将Scala @suspendable方法转换为未来

时间:2011-09-03 16:12:10

标签: scala continuations

假设我有睡眠功能:

def sleep(delay:Int) : Unit @suspendable = {
  ....
}

是否可以使用函数future来创建可以同步等待的睡眠函数的异步版本。

def future(targetFunc: (Int => Unit @suspendable)) : (Int => Future) = {
    ....
}

class Future {
  def await : Unit @suspendable = {
     ....
  }
}

你应该可以做这样的事情:

reset {
  val sleepAsync = future(sleep)
  val future1 = sleepAsync(2000)
  val future2 = sleepAsync(3000)
  future1.await
  future2.await
  /* finishes after a delay of 3000 */
}

对sleepAsync的两次调用应该会立即返回,对Future#await的两次调用应该会阻塞。当然它们都会在重置结束时掉线,后面的代码负责在延迟后调用延续。

否则有另一种方法可以并行运行两个@suspendable函数并等待它们完成?

我有一个可编辑的要点,其中包含我想做的骨架:https://gist.github.com/1191381

2 个答案:

答案 0 :(得分:2)

object Forks {

  import scala.util.continuations._

  case class Forker(forks: Vector[() => Unit @suspendable]) {
    def ~(block: => Unit @suspendable): Forker = Forker(forks :+ (() => block))
    def joinIf(pred: Int => Boolean): Unit @suspendable = shift { k: (Unit => Unit) =>
      val counter = new java.util.concurrent.atomic.AtomicInteger(forks.size)
      forks foreach { f =>
        reset {
          f()
          if (pred(counter.decrementAndGet)) k()
        }
      }
    }
    def joinAll() = joinIf(_ == 0)
    def joinAny() = joinIf(_ == forks.size - 1)
  }

  def fork(block: => Unit @suspendable): Forker = Forker(Vector(() => block))
}

使用fork(),我们现在可以等待许多“suspendables”。使用〜()将suspendables链接在一起。使用joinAll()等待所有suspendables和joinAny()等待只有一个。使用joinIf()来自定义连接策略。

object Tests extends App {

  import java.util.{Timer, TimerTask}
  import scala.util.continuations._

  implicit val timer = new Timer

  def sleep(ms: Int)(implicit timer: Timer): Unit @suspendable = {
    shift { k: (Unit => Unit) =>
      timer.schedule(new TimerTask {
        def run = k()
      }, ms)
    }
  }

  import Forks._

  reset {
    fork {
      println("sleeping for 2000 ms")
      sleep(2000)
      println("slept for 2000 ms")
    } ~ {
      println("sleeping for 4000 ms")
      sleep(4000)
      println("slept for 4000 ms")
    } joinAll()
    println("and we are done")
  }
  println("outside reset")
  readLine
  timer.cancel
}

这是输出。程序从时间T开始:

sleeping for 2000 ms
sleeping for 4000 ms
outside reset         <<<<<< T + 0 second
slept for 2000 ms     <<<<<< T + 2 seconds
slept for 4000 ms     <<<<<< T + 4 seconds
and we are done       <<<<<< T + 4 seconds

答案 1 :(得分:1)

我不确定我是否完全理解这个问题,但这是一次尝试:

import scala.util.continuations._

class Future(thread: Thread) {
  def await = thread.join
}

object Future {

  def sleep(delay: Long) = Thread.sleep(delay)

  def future[A,B](f: A => B) = (a: A) => shift { k: (Future => Unit) =>
    val thread = new Thread { override def run() { f(a) } }
    thread.start()

    k(new Future(thread))
  }

  def main(args:Array[String]) = reset {
    val sleepAsync = future(sleep)
    val future1 = sleepAsync(2000) // returns right away
    val future2 = sleepAsync(3000) // returns right away
    future1.await // returns after two seconds
    future2.await // returns after an additional one second
    // finished after a total delay of three seconds
  }
}

此处,Future实例仅仅是Thread上的句柄,因此您可以使用其join方法进行阻止,直至完成。

future函数采用类型A => B的函数,并返回一个函数,当提供A时,该函数将启动线程以运行“futured”函数,并且将其包装在Future中,然后将其注入连续符,从而将其分配给val future1

这是否与您的目标相近?