Scala: return a parametrized instance - scala

Question

I cannot make work the following code:
object Factory {
def apply[U <: Cda](type: MyType.Value): MyUtilTrait[U] = {
type match {
case MyType.Value.one => MyOneUtilCustom
case MyType.Value.two => MyTwoUtilCustom
}
}
}
=> Expression of type Factory.MyType doesn't conform to expected type MyUtilTrait[U]
trait MyUtilTrait[T <: Cda] {}
object MyOneUtilCustom extends MyUtilTrait[CdaOneCustom] { }
object MyTwoUtilCustom extends MyUtilTrait[CdaTwoCustom] { }
case class CdaOneCustom(...) extends Cda {}
case class CdaTwoCustom(...) extends Cda {}
abstract class Cda(...) {}
object MyType extends Enumeration {
val one, two = Value
}
With the apply, I am supposed to return a MyUtilTrait parametrized with a subtype of Cda, so what's wrong?

Answer 1

It's almost completely impossible to implement this apply method with such a signature [1], because someone could come along, define
class Unobtanium extends Cda {
// implement all `Cda` methods by `???`
}
and then invoke
Factory.apply[Unobtanium](MyType.one)
How is a factory supposed to create a MyUtilTrait[Unobtanium], if it knows nothing about Unobtanium, and it is the first time it sees this strange type?
Use existential type instead:
abstract class Cda {}
case class CdaOneCustom() extends Cda {}
case class CdaTwoCustom() extends Cda {}
trait MyUtilTrait[T <: Cda] {}
object MyOneUtilCustom extends MyUtilTrait[CdaOneCustom] { }
object MyTwoUtilCustom extends MyUtilTrait[CdaTwoCustom] { }
object MyType extends Enumeration {
val one, two = Value
}
object Factory {
def apply(typ: MyType.Value): MyUtilTrait[_] = {
import MyType._
typ match {
case `one` => MyOneUtilCustom
case `two` => MyTwoUtilCustom
}
}
}
[1] Unless your MyUtilTrait[X] is something trivial, like Nil (which is a List[X] for any X), or Consumer[Any] of some sort, which does not actually care about the type parameter.