Scala higher kinded type variance - scala

Question

I am dipping my toes in higher kinded types, exploring a very basic Scala example:
trait Mappable[F[_]] {
def map[A, B](fa: F[A])(f: A => B): F[B]
}
object Mappable {
implicit object MappableOption extends Mappable[Option] {
def map[A, B](fa: Option[A])(f: A => B): Option[B] = fa.map(f)
}
implicit object MappableSeq extends Mappable[Seq] {
def map[A, B](fa: Seq[A])(f: A => B): Seq[B] = fa.map(f)
}
}
def bananaTuple[F[_], T](f: F[T])(implicit F: Mappable[F]): F[(String, T)] =
F.map(f)(("banana", _))
This works:
bananaTuple(Option(42)) // Some((banana,42))
bananaTuple(Seq(42)) // List((banana,42))
But this does not compile:
bananaTuple(Some(42))
bananaTuple(List(42))
The compile errors I get:
could not find implicit value for parameter F: ch.netzwerg.hkt.HigherKindedTypes.Mappable[Some] bananaTuple(Some(42))
not enough arguments for method bananaTuple: (implicit F: ch.netzwerg.hkt.HigherKindedTypes.Mappable[Some])Some[(String, Int)]. Unspecified value parameter F. bananaTuple(Some(42))
How can I bring variance into the game?

Answer 1

We can make this work with a little more parameteric polymorphism:
object MappableExample {
trait Mappable[F[_]] {
type Res[_]
def map[A, B](f: A => B)(c: F[A]): Res[B]
}
implicit def seqMappable[C[X] <: Seq[X]] = new Mappable[C] {
type Res[X] = Seq[X]
override def map[A, B](f:A => B)(c: C[A]): Seq[B] = c.map(f)
}
implicit def optionMappable[C[X] <: Option[X]]: Mappable[C] = new Mappable[C] {
type Res[X] = Option[X]
override def map[A, B](f: A => B)(c: C[A]): Option[B] = c.map(f)
}
def map[A, B, C[_]](xs: C[A])(f: A => B)(implicit mappable: Mappable[C]): mappable.Res[B] = {
mappable.map(f)(xs)
}
def main(args: Array[String]): Unit = {
println(map(List(1,2,3))(("banana", _)))
println(map(Some(1))(("banana", _)))
}
}
Yields:
List((banana,1), (banana,2), (banana,3))
Some((banana,1))
The compiler now infers Some as Mappable[Some]#Res[Int] and Mappable[List]#Res[Int] which is quite ugly. One would expect the compiler to actually be able to infer the right type without needing for any co/contravariance on the Mappable trait, which we can't do since we're using it in an invariant position.

Answer 2

Subtype polymorphism allows us to pass values of a certain type or any of its subtypes to a method. If a method takes a value of type Fruit, we can also pass an Apple inside (an apple is a fruit after all). So if you want to be able to pass a Mappable.MappableOption to your bananaTuple method, you have to make that MappableOption a subtype of MappableSome (since the type of your first parameter of bananaTuple dictates the implicit one). This means that you want your Mappable contravariant (if Some <: Option, then Mappable[Some] >: Mappable[Option]).
But you cannot have Mappable[F[_]] contravariant in F because F appears in covariant position of map (as a function parameter). Note that F also appears in contravariant position of map (as a return value).
If you manage to make Mappable[F[_]] contravariant in F, it should work, but I'm not sure if making it contravariant makes sense. That is, if you want a subtype relationship such as e.g. Apple <: Fruit to result in Mappable[Apple] >: Mappable[Fruit] (this would not compile since Apple and Fruit are not type constructors, but I'm just using simple types to make a point here).
Making a type contravariant in its type and solving the problem of contravariant type appearing in covariant position is a common problem and perhaps it's better if you search for it elsewhere (here is one example). I still think that it's better to provide an implicit object for every type you want to use, that is, to provide separate implicit objects for e.g. Seq and List.