Composable using scalaz Arrow?

Question

I have two functions.

  def process(date: DateTime, invoice: Invoice, user: User, reference: Reference) : (Action, Iterable[Billable])

  def applyDiscount(billable: Billable) : Billable

How can I compose these so that I have a single function of (DateTime, Invoice, User, Reference) => (Action, Iterable[Billable])

Here is the poor mans way of what I want

  def buildFromInvoice(user: User, order: Invoice, placementDate: DateTime, reference: Reference) = {
    val ab = billableBuilder.fromInvoice(user, order, placementDate, reference)
    (ab._1, ab._2.map(applyDiscount(_))
  }

Answer 1

What you have (simplifying) is:

val f: A => (B, M[C]) //M is a Functor
val g: C => C

I can think of a few ways of doing this. I think my preference is:

(a: A) => g.lift[M].second apply f(a)

Or also:

(a: A) => f(a) :-> g.lift[M]  

However, there is possibly a pointfree way - although not necessarily so, of course

• lift is a method on Function1W which lifts the function into the realm of the functor M
• second is a method on MAB which applies the function down the right-hand-side of a Bifunctor
• :-> is a method available to Bifunctors denoting the application of a function on the rhs.

EDIT - missingfaktor appears to be correct in saying f andThen g.lift[M].second works:

scala> import scalaz._; import Scalaz._
import scalaz._
import Scalaz._

scala> case class A(); case class B(); case class C()
defined class A
defined class B
defined class C

scala> lazy val f: A => (B, List[C]) = sys.error("")
f: A => (B, List[C]) = <lazy>

scala> lazy val g: C => C = sys.error("")
g: C => C = <lazy>

Pointfree:

scala> lazy val h = f andThen g.lift[List].second
h: A => (B, List[C]) = <lazy>