重复eval T => scala.concurrent.Future [T]到一个进程[?,T]
发布时间:2020-12-16 18:05:10 所属栏目:安全 来源:网络整理
导读:我有一个函数get:T = scala.concurrent.Future [T] 我想迭代它像: val futs: Iterator[Future[T]] = Iterator.iterate(get(init)){ _.flatMap(prev = get(prev)) } 但Iterator的类型是Future [T],处理这个迭代器并不容易. How could I transfer that to Pr
我有一个函数get:T =>
scala.concurrent.Future [T]
我想迭代它像: val futs: Iterator[Future[T]] = Iterator.iterate(get(init)){ _.flatMap(prev => get(prev)) } 但Iterator的类型是Future [T],处理这个迭代器并不容易.
(也许T => Future [T]作为上下文类型F). 解决方法
不是超级好的解决方案,但有效
import scala.concurrent.ExecutionContext.Implicits.global import scala.concurrent.{Future => SFuture} import scala.language.implicitConversions import scalaz.concurrent.Task import scalaz.stream._ implicit class Transformer[+T](fut: => SFuture[T]) { def toTask(implicit ec: scala.concurrent.ExecutionContext): Task[T] = { import scala.util.{Success,Failure} import scalaz.syntax.either._ Task.async { register => fut.onComplete { case Success(v) => register(v.right) case Failure(ex) => register(ex.left) } } } } val init: Int = 0 def f(i: Int): SFuture[Int] = SFuture(i + 1) val p = Process.repeatEval[Task,Int] { var prev = init f(prev).toTask.map(next => {prev = next; next}) } println(p.take(10).runLog.run) (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |