Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

What's the relation of fold on Option, Either etc and fold on Traversable?

Scalaz provides a method named fold for various ADTs such as Boolean, Option[_], Validation[_, _], Either[_, _] etc. This method basically takes functions corresponding to all possible cases for that given ADT. In other words, a pattern match shown below:

x match {
  case Case1(a, b, c) => f(a, b, c)
  case Case2(a, b) => g(a, b)
  .
  .
  case CaseN => z
}

is equivalent to:

x.fold(f, g, ..., z)

Some examples:

scala> (9 == 8).fold("foo", "bar")
res0: java.lang.String = bar

scala> 5.some.fold(2 *, 2)
res1: Int = 10

scala> 5.left[String].fold(2 +, "[" +)
res2: Any = 7

scala> 5.fail[String].fold(2 +, "[" +)
res6: Any = 7

At the same time, there is an operation with the same name for the Traversable[_] types, which traverses over the collection performing certain operation on its elements, and accumulating the result value. For example,

scala> List(2, 90, 11).foldLeft("Contents: ")(_ + _.toString + " ")
res9: java.lang.String = "Contents: 2 90 11 "

scala> List(2, 90, 11).fold(0)(_ + _)
res10: Int = 103

scala> List(2, 90, 11).fold(1)(_ * _)
res11: Int = 1980

Why are these two operations identified with the same name - fold/catamorphism? I fail to see any similarities/relation between the two. What am I missing?

like image 877
missingfaktor Avatar asked Dec 16 '11 20:12

missingfaktor


2 Answers

I think the problem you are having is that you see these things based on their implementation, not their types. Consider this simple representation of types:

List[A] = Nil 
        | Cons head: A tail: List[A]

Option[A] = None
          | Some el: A

Now, let's consider Option's fold:

fold[B] = (noneCase: => B, someCase: A => B) => B

So, on Option, it reduces every possible case to some value in B, and return that. Now, let's see the same thing for List:

fold[B] = (nilCase: => B, consCase: (A, List[A]) => B) => B

Note, however, that we have a recursive call there, on List[A]. We have to fold that somehow, but we know fold[B] on a List[A] will always return B, so we can rewrite it like this:

fold[B] = (nilCase: => B, consCase: (A, B) => B) => B

In other words, we replaced List[A] by B, because folding it will always return a B, given the type signature of fold. Now, let's see Scala's (use case) type signature for foldRight:

foldRight[B](z: B)(f: (A, B) ⇒ B): B

Say, does that remind you of something?

like image 166
Daniel C. Sobral Avatar answered Nov 03 '22 20:11

Daniel C. Sobral


If you think of "folding" as "condensing all the values in a container through an operation, with a seed value", and you think of an Option as a container that can can have at most one value, then this starts to make sense.

In fact, foldLeft has the same signature and gives you exactly the same results if you use it on an empty list vs None, and on a list with only one element vs Some:

scala> val opt : Option[Int] = Some(10)
opt: Option[Int] = Some(10)

scala> val lst : List[Int] = List(10)
lst: List[Int] = List(10)

scala> opt.foldLeft(1)((a, b) => a + b)
res11: Int = 11

scala> lst.foldLeft(1)((a, b) => a + b)
res12: Int = 11

fold is also defined on both List and Option in the Scala standard library, with the same signature (I believe they both inherit it from a trait, in fact). And again, you get the same results on a singleton list as on Some:

scala> opt.fold(1)((a, b) => a * b)
res25: Int = 10

scala> lst.fold(1)((a, b) => a * b)
res26: Int = 10

I'm not 100% sure about the fold from Scalaz on Option/Either/etc, you raise a good point there. It seems to have quite a different signature and operation from the "folding" I'm used to.

like image 36
Ben Avatar answered Nov 03 '22 20:11

Ben