CODEWITHSUNDEEP
scalarecursion
Det
Det

Reputation: 3710

Scala Recursion as per list contents

I'm attempting to do a method of an object x times, as per the contents of, say:

val obstacles = Vector[Obstacle](list...)

What I'm doing currently is just suffixing all out (where locations are of type, say Location):

val background: Pic = rectangle
background.place(obstacle1.shape, obstacle1.location)
          .place(obstacle2.shape, obstacle2.location)
          .place(shape3Pic, new Location(location3))
          .place(shape4Pic, new Location(location4))...

..which I tried to do with map:

obstacles.map(o => background.place(o.shape, o.location))

..however this then of course only returns a Vector[Obstacle], not the original background: Pic with the "additions", and will not work with the shape3Pic: Pic and shape4Pic: Pic, as they are not Obstacle.

Apparently this can be done in recursion?

Upvotes: 0

Views: 65

Answers (2)

cheseaux
cheseaux

Reputation: 5325

You could also do it without recursion using foldLeft

shapes.foldLeft(background)((r,c) => r.place(c))

Or even shorter

shapes.foldLeft(background)(_ place _)

Upvotes: 3

Feyyaz
Feyyaz

Reputation: 3216

Assuming Background.place takes only Shape object, it could be as follows:

trait Shape { }

trait Background {
  def place(shape: Shape): Background
}

@tailrec
def placeRecursive(shapes: Vector[Shape], background: Background): Background = {
  shapes match {
    case shape +: rest =>
      placeRecursive(rest, background.place(shape))
    case _ =>
      background
  }
}

Upvotes: 1

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