Scala - avoid use mutable variables

Question

I have a function that perform a calc but i'm using a var to receive the value of a recursive function and i would like to avoid mutable variables.

def scoreNode(node:Option[Tree], score:Double = 0, depth:Int = 0):Double = {
  node.map(n => {
    var points = score
    n.children.filter(n => n.valid == Some(true)).foreach(h => {
      points = scoreNode(Some(h), 10, depth+1)
    })
    points
  }).getOrElse(score)
}

How can i rewrite this piece of code without a mutable variable? I've tried

You'll often find that any time you accept an argument of Option and immediately map on it, that your code gets easier to manage if you change the type signature to accept the non-optional type and map at the call site. That would remove some of the noise here and make it easier for you to see the behavior alone — Daenyth
– Daenyth, Commented Nov 5, 2016 at 12:57
What do you mean? I come from OO and FP seens obscure to me. Could you give an example? — user866364
– user866364, Commented Nov 5, 2016 at 12:59
Don't write scoreNode(node: Option[Tree], ...) = node.map(...) instead write scoreNode(node: Tree, ...) = ... and call it by maybeNode.map(scoreNode).getOrElse(defaultScore) — Daenyth
– Daenyth, Commented Nov 5, 2016 at 13:06
have you considered using pattern matching? Recursion and Pattern matching goes well with Scala way of doing stuff and avoiding mutation — DevZer0
– DevZer0, Commented Nov 5, 2016 at 13:36

Rok Kralj · Accepted Answer · 2016-11-05 12:52:28Z

1

What you are essentially doing is summing something over all the nodes in a tree. Try to write a more idiomatic code, like this.

def scoreNode(node:Option[Tree], depth:Int = 0):Double =
    (for {
        n <- node
        h <- n.children
        if h.valid == Some(true)
        res = scoreNode(Some(h), depth + 1) + scala.math.pow(0.8, depth)
    } yield res).sum

I do not guarantee this works completely. It is your homework to make it right.

answered Nov 5, 2016 at 12:52

Rok Kralj

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Comments

Daniel Shin · Accepted Answer · 2016-11-05 12:35:40Z

0

You can use fold:

def scoreNode(node:Option[Tree], score:Double = 0, depth:Int = 0):Double = 
  node
    .map(_.children.filter(n => n.valid == Some(true)).fold(score)((acc, h) => scoreNode(Some(h), acc + scala.math.pow(0.8, depth), depth + 1)))
    .getOrElse(score)

answered Nov 5, 2016 at 12:35

Daniel Shin

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3 Comments

Daniel Shin Over a year ago

I coudn't check it to compile because Tree type is not given. Probably had some typo somewhere. Can you provide type for Tree?

user866364 Over a year ago

is possible to avoid cast on acc and h? acc.asInstanceOf[Double]?

Daniel Shin Over a year ago

You can refine type for fold like this: fold[Double](...

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