Scala functional programming gymnastics

Question

I am trying to do the following in as little code as possible and as functionally as possible:

def restrict(floor : Option[Double], cap : Option[Double], amt : D         


        

Answer 1

Edit 2:

While thinking about the cataX method, I figured out that cataX is nothing else than a plain and simple fold. Using that, we can get a pure scala solution without any additional libraries.

So, here it is:

( (amt /: floor)(_ max _) /: cap)(_ min _)

which is the same as

cap.foldLeft( floor.foldLeft(amt)(_ max _) )(_ min _)

(not that this is necessarily easier to understand).

I think you can’t have it any shorter than that.

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For better or worse, we can also solve it using scalaz:

floor.map(amt max).getOrElse(amt) |> (m => cap.map(m min).getOrElse(m))

or even:

floor.cata(amt max, amt) |> (m => cap.cata(m min, m))

As a ‘normal’ Scala programmer, one might not know about the special Scalaz operators and methods used (|> and Option.cata). They work as follows:

value |> function translates to function(value) and thus amt |> (m => v fun m) is equal to v fun amt.

opt.cata(fun, v) translates to

opt match {
  case Some(value) => fun(value) 
  case None => v
}

or opt.map(fun).getOrElse(v).

See the Scalaz definitions for cata and |>.

A more symmetric solution would be:

amt |> (m => floor.cata(m max, m)) |> (m => cap.cata(m min, m))

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Edit: Sorry, it’s getting weird now, but I wanted to have a point-free version as well. The new cataX is curried. The first parameter takes a binary function; the second is a value.

class CataOption[T](o: Option[T]) {
  def cataX(fun: ((T, T) => T))(v: T) = o.cata(m => fun(m, v), v)
}
implicit def option2CataOption[T](o: Option[T]) = new CataOption[T](o)

If o matches Some we return the function with the value of o and the second parameter applied, if o matches None we only return the second parameter.

And here we go:

amt |> floor.cataX(_ max _) |> cap.cataX(_ min _)

Maybe they already have this in Scalaz…?