Using scalaz.Unapply

by Stephen Compall on Sep 11, 2013


Once you’ve started really taking advantage of Scalaz’s typeclasses for generic programming, you might have noticed a need to write typelambdas to use some of your neat abstractions, or use syntax like traverse or kleisli with a strangely-shaped type as an argument. Here’s a simple generalization, a List-based traverse.

import scalaz.Applicative, scalaz.syntax.applicative._

def sequenceList[F[_]: Applicative, A](xs: List[F[A]]): F[List[A]] =
  xs.foldRight(List.empty[A].point[F])((a, b) => ^(a, b)(_ :: _))

This works fine for a while.

scala> import scalaz.std.option._
import scalaz.std.option._

scala> sequenceList(List(some(1),some(2)))
res1: Option[List[Int]] = Some(List(1, 2))

scala> sequenceList(List(some(1),none))
res2: Option[List[Int]] = None

The problem

The type of the input in the above example, List[Option[Int]], can be neatly destructured into the F and A type params needed by sequenceList. It has the “shape” F[x], so F can be picked out by Scala easily.

Consider something else with a convenient Applicative instance, though.

scala> import scalaz.\/
import scalaz.$bslash$div

scala> sequenceList(List(\/.right(42), \/.left(NonEmptyList("oops"))))
<console>:23: error: no type parameters for method 
  sequenceList: (xs: List[F[A]])(implicit evidence$1: scalaz.Applicative[F])F[List[A]]
  exist so that it can be applied to arguments
 --- because ---
argument expression's type is not compatible with formal parameter type;
 found   : List[scalaz.\/[scalaz.NonEmptyList[String],Int]]
 required: List[?F]

              sequenceList(List(\/.right(42), \/.left(NonEmptyList("oops"))))

Here, ?F meaning it couldn’t figure out that you meant ({type λ[α] = NonEmptyList[String] \/ α})#λ.

scala> sequenceList[({type λ[α] = NonEmptyList[String] \/ α})#λ, Int
                  ](List(\/.right(42), \/.left(NonEmptyList("oops"))))
res5: scalaz.\/[scalaz.NonEmptyList[String],List[Int]] =

The problem is that NonEmptyList[String] \/ Int has the shape F[A, B], with F of kind * -> * -> * after a fashion, whereas the F it wants must have kind * -> *, and Scala kinds aren’t curried at all.

Finding an Unapply instance

Unapply, though, does have implicit instances matching the F[A, B] shape, unapplyMAB1 and unapplyMAB2, in its companion so effectively always visible. What’s special about them is that their type parameters match the “shape” you’re working with, F[A, B].

You should look at their source to follow along.

Let’s see if one of them works. For implicit resolution to finish, it’s important that exactly one of them works.

scala> import scalaz.Unapply
import scalaz.Unapply

scala> Unapply.unapplyMAB1[Applicative, \/, NonEmptyList[String], Int]
<console>:23: error: could not find implicit value for parameter
TC0: scalaz.Applicative[[α]scalaz.\/[α,Int]]
              Unapply.unapplyMAB1[Applicative, \/, NonEmptyList[String], Int]

scala> Unapply.unapplyMAB2[Applicative, \/, NonEmptyList[String], Int]
res7: scalaz.Unapply[scalaz.Applicative,
        type M[X] = scalaz.\/[scalaz.NonEmptyList[String],X];
        type A = Int
      } = scalaz.Unapply_0$$anon$13@5402af61

Here, the type res7.M represents the typelambda being passed to sequenceList. You can see that work.

scala> sequenceList[res7.M, res7.A](
                   List(\/.right(42), \/.left(NonEmptyList("oops"))))
res8: res7.M[List[res7.A]] = -\/(NonEmptyList(oops))

scala> res8 : NonEmptyList[String] \/ List[Int]
res9: scalaz.\/[scalaz.NonEmptyList[String],List[Int]] =

The res8 conformance test shows that Scala can still reduce the path-dependent res7.M and res7.A types at this level, outside sequenceList.

Searching for the right shape

Implicit resolution can pick the call to unapplyMAB2 partly because it can pick all of its type parameters without weird typelambda structures. But in Scalaz, we use typeclasses to guide its choice.

Why didn’t unapplyMAB1 work? In this case, you can trust scalac to say exactly the right thing: it looked for Applicative[[α]scalaz.\/[α,Int]], and didn’t find one. Sure enough, \/ being right-biased means we don’t offer that instance.

Incidentally, if you were to introduce that instance, you’d break code relying on right-biased Unapply resolution to work.

unapplyMAB2 needs evidence of TC[({type λ[α] = M0[A0, α]})#λ]. But that’s okay, because we have that, where TC=Applicative, M0=\/, and A0=NonEmptyList[String]!

scala> Applicative[({type λ[α] = \/[NonEmptyList[String], α]})#λ]
res10: scalaz.Applicative[[α]scalaz.\/[scalaz.NonEmptyList[String],α]]
         = scalaz.DisjunctionInstances2$$anon$1@2f658816

Scala doesn’t need to figure out any typelambda itself for this to work; we did everything by putting the typelambda right into unapplyMAB2’s evidence requirement, so it just has to find the conforming implicit value.

Using Unapply generically

Now you can write a sequenceList wrapper that works for \/ and many other shapes, including user-provided shapes in the form of new Unapply implicit instances. If you’re using Scala 2.9 (still?!) you need to add -Ydependent-method-types to scalacOptions to write this function.

def sequenceListU[FA](xs: List[FA])
                     (implicit U: Unapply[Applicative, FA]
                     ): U.M[List[U.A]] =

Instead of xs being List[F[A]], it’s List[FA], and that’s destructured into U.M and U.A. The latter are path-dependent types on U, the conventional name of the Unapply parameter. We have also followed the convention of naming the Unapply-taking variant function ending with a U.

And that works great!

scala> sequenceListU(List(\/.right(42), \/.left(NonEmptyList("oops"))))
res11: scalaz.\/[scalaz.NonEmptyList[String],List[Int]] =

Of course, there’s that strange-looking function body to consider, still.

Using the U evidence

The type equalities of the original U.M and U.A to the original types can be seen where res8 is refined to res9 above. But only the caller of the function knows those equalities, because it produced and supplied the unapplyMAB2 call, which has a structural type containing those equalities.

The body of sequenceListU doesn’t know those things. In particular, it still can’t pick type parameters to pass to sequenceList without a little help.

The leibniz member is a reified type equality of FA === U.M[U.A], meaning those are the same at the type level, even though Scala can’t see it in this context. It represents genuine evidence that those two types are equal, and is much more powerful than scala-library’s own =:=. We’re using the core Leibniz operator, subst, directly to prove that, as a consequence of that type equality, List[FA] === List[U.M[U.A]] is also a type equality, and that therefore this [constant-time] coercion is valid. This lifting is applicable in all contexts, not just covariant ones like List’s. Take a look at the full API for more, though you’ll typically just need to come up with the right type parameter for subst.

You can’t ask for an Unapply and also ask for an Applicative[U.M]; Scala won’t allow it. So, because we needed to resolve the typeclass anyway to find the Unapply implicit to use, we just cart it along with the U and give it to the function, which almost always needs to use it anyway. Because it’s not implicitly available, you usually need to grab it, U.TC, and use it directly.

Using in scalaz.syntax

map comes from functor syntax; it’s not a method on Function1. So how come this works?

scala> import scalaz.std.function._
import scalaz.std.function._

scala> ((_:Int) + 42) map (_ * 33)
res13: Int => Int = <function1>

scala> res13(1)
res14: Int = 1419

When you import syntax, as Functor syntax was imported with scalaz.syntax.applicative._ above, you get at least two conversions: the plain one, like ToFunctorOps[F[_],A], which works if you have the right shape, and the fancy one, ToFunctorOpsUnapply[FA], which uses an Unapply to effectively invoke ToFunctorOps as in the above. The latter is lower-priority, so Scala will pick the former if the value has the F[A] shape.

That gives access to all the methods in FunctorOps, and other ops classes, with only one special U-taking method. If you have several functions operating on the same value type, or you can make that type similar with Leibnizian equality as implicit arguments to your methods, I suggest grouping them in this way, too, to cut down on boilerplate.

Provide both anyway

We sometimes get asked “why not just provide the Unapply version of the function or ops?”

We do it, and suggest it for your own code, despite the confusion, because it’s easier to work with real type equalities than with Leibnizian equality, which you can do in your “real” function implementation, and as seen in res8 above, the path-dependent type resolution can leave funny artifacts in the inferred result. Here’s an extreme example from an earlier demonstration.

scala> val itt = IdentityT lift it
itt: IdentityT[scalaz.Unapply[scalaz.Monad,IdentityT[scalaz.Unapply[scalaz.Monad,Identity[Int]]
                                                       {type M[X] = Identity[X]; type A = Int}#M,
                 {type M[X] = IdentityT[scalaz.Unapply[scalaz.Monad,Identity[Int]]
                                          {type M[X] = Identity[X]; type A = Int}#M,
                  type A = Int}#M,
  = IdentityT(IdentityT(Identity(42)))


Jason Zaugg implemented Scalaz Unapply, based on ideas from Miles Sabin and Paul Chiusano.

Leibnizian equality was implemented for Scalaz by Edward Kmett.

Lars Hupel’s talk (slides, video) on the features in the then-upcoming Scalaz 7 at nescala 2013, including Unapply, gave me the missing “guided by typeclasses” detail, inspiring me to tell more people about the whole thing at the conference, and then, much later, write it down here.

This article was tested with Scala 2.10.2 & Scalaz 7.0.3.