Question

The code below works but pattern matching on Option b doesn’t look elegant.
Is there a way to avoid it while keeping the same semantics?

object A {
  def apply(b: B): ValidationNEL[String, A] = ...
}

case class C(i: Int, a: Option[A])

object C {
  def apply(i: Int, b: Option[B]): ValidationNEL[String, C] = b match {
    case None => Success(C(i, None))
    case Some(sb) => A(sb).map(bb => C(i, Some(bb)))
  }     
}

Answer 1

Using scalaz, you can fold over Option

b fold ( sb => A(sb) map (bb => C(i, some(bb))), C(i, none).success)

There’s probably a way of simplifying the sb => A(sb) map (bb => C(i, some(bb))) using pointfree style but this is often ugly in scala:

Setup

scala> :paste
// Entering paste mode (ctrl-D to finish)

object A {
  def apply(b: B): ValidationNEL[String, A] = sys.error("")
}
case class A()
case class B()
case class C(i: Int, a: Option[A])

// Exiting paste mode, now interpreting.

defined module A
defined class A
defined class B
defined class C

First impl

scala> def apply(i: Int, b: Option[B]): ValidationNEL[String, C] =
     | b fold ( sb => A(sb) map (bb => C(i, some(bb))), C(i, none).success)
apply: (i: Int, b: Option[B])scalaz.Scalaz.ValidationNEL[String,C]

Second Impl

If you declare first-class functions, you stand a better chance of composition. For example:

object A { val fromB: B => ValidationNEL[String, A] = _ => sys.error("") }
object C { val fromA: Int => A => C = i => a => C(i, some(a)) }

defined module A
defined module C

Then

scala> def apply(i: Int, b: Option[B]): ValidationNEL[String, C] =
     | b fold (A.fromB andThen (_ map C.fromA(i)), C(i, none).success)
apply: (i: Int, b: Option[B])scalaz.Scalaz.ValidationNEL[String,C]