scala – 什么是适当的monad或序列理解来映射和携带状态？

我需要正确的代码习语来评估一系列表达式以获取其值的序列,并在评估发生时将状态从一个评估者传播到下一个评估者.我想要一个功能性的编程成语.

这不是折叠,因为结果像地图一样出现.这不是一个地图,因为国家的支柱.

我所用的是这个代码,我用来试图弄清楚这一点.熊先生几条测试台：

// test rig
class MonadLearning extends JUnit3Suite {

  val d = List("1","2","3") // some expressions to evaluate. 

  type ResType = Int 
  case class State(i : ResType) // trivial state for experiment purposes
  val initialState = State(0)

// my stub/dummy "eval" function...obviously the real one will be...real.
  def computeResultAndNewState(s : String,st : State) : (ResType,State) = {
    val State(i) = st
    val res = s.toInt + i
    val newStateInt = i + 1
    (res,State(newStateInt))
  }

我目前的解决方案使用更新的var,因为地图的正文被评估：

def testTheVarWay() {
    var state = initialState
    val r = d.map {
      s =>
        {
          val (result,newState) = computeResultAndNewState(s,state)
          state = newState
          result
        }
    }
    println(r)
    println(state)
  }

我有我认为使用foldLeft的不可接受的解决方案,这就是我所说的“把它当成你的折叠”成语：

def testTheFoldWay() {

// This startFold thing,requires explicit type. That alone makes it muddy.
val startFold : (List[ResType],State) = (Nil,initialState)
val (r,state) = d.foldLeft(startFold) {
  case ((tail,st),s) => {
    val (r,ns) = computeResultAndNewState(s,st)
    (tail :+ r,ns) // we want a constant-time append here,not O(N). Or could Cons on front and reverse later
  }
}

println(r)
println(state)

}

我也有几个递归的变化(这是显而易见的,但也不清楚或有很好的动机),一个使用几乎可以容忍的流：

def testTheStreamsWay() {
  lazy val states = initialState #:: resultStates // there are states
  lazy val args = d.toStream // there are arguments
  lazy val argPairs = args zip states // put them together
  lazy val resPairs : Stream[(ResType,State)] = argPairs.map{ case (d1,s1) => computeResultAndNewState(d1,s1) } // map across them
  lazy val (results,resultStates) = myUnzip(resPairs)// Note .unzip causes infinite loop. Had to write my own.

  lazy val r = results.toList
  lazy val finalState = resultStates.last

  println(r)
  println(finalState)
}

但是,我无法确定任何像上面原来的“var”解决方案那样紧凑或清晰的东西,我愿意和他一起生活,但是我认为吃/喝/睡眠monad成语的人会说..使用这个…(希望！)