FoldLeft via FoldRight in Scala again

Question

I know, this question has been asked already. But I have not understood any of the answers. I think I need a more graphic explanation. I can't understand how to "bridge" FoldLeft with FoldRight. I don't care if the anwer is not in Functional Programming in Scala. Thabk you very much in advance.

Answer 1

Just check how those are implemented:

  def foldLeft[B](z: B)(op: (B, A) => B): B = {
    var result = z
    this foreach (x => result = op(result, x))
    result
  }

  def foldRight[B](z: B)(op: (A, B) => B): B =
    reversed.foldLeft(z)((x, y) => op(y, x))

foldLeft traverses collection from left to right applying op to the result and current element, while foldRight traverses reversed collection (ie from right to left).

When op is symmetric and transitive foldLeft and foldRight are equivalent, for example:

List(1,2,3).foldLeft(0)(_ + _)
List(1,2,3).foldRight(0)(_ + _)

Result:

res0: Int = 6
res1: Int = 6

But otherwise foldLeft and foldRight may produce different results:

List(1,2,3).foldLeft(List[Int]()){case (list, el) => list :+ el }
List(1,2,3).foldRight(List[Int]()){case (el, list) => list :+ el }

Result:

res2: List[Int] = List(1, 2, 3)
res3: List[Int] = List(3, 2, 1)