How to define “type disjunction” (union types)?

Question

One way that has been suggested to deal with double definitions of overloaded methods is to replace overloading with pattern matching:

object Bar {
   def fo         


        

Answer 1

I have sort of stumbled on a relatively clean implementation of n-ary union types by combining the notion of type lists with a simplification of Miles Sabin's work in this area, which someone mentions in another answer.

Given type ¬[-A] which is contravariant on A, by definition given A <: B we can write ¬[B] <: ¬[A], inverting the ordering of types.

Given types A, B, and X, we want to express X <: A || X <: B. Applying contravariance, we get ¬[A] <: ¬[X] || ¬[B] <: ¬[X]. This can in turn be expressed as ¬[A] with ¬[B] <: ¬[X] in which one of A or B must be a supertype of X or X itself (think about function arguments).

object Union {
  import scala.language.higherKinds

  sealed trait ¬[-A]

  sealed trait TSet {
    type Compound[A]
    type Map[F[_]] <: TSet
  }

  sealed trait ∅ extends TSet {
    type Compound[A] = A
    type Map[F[_]] = ∅ 
  }

  // Note that this type is left-associative for the sake of concision.
  sealed trait ∨[T <: TSet, H] extends TSet {
    // Given a type of the form `∅ ∨ A ∨ B ∨ ...` and parameter `X`, we want to produce the type
    // `¬[A] with ¬[B] with ... <:< ¬[X]`.
    type Member[X] = T#Map[¬]#Compound[¬[H]] <:< ¬[X]

    // This could be generalized as a fold, but for concision we leave it as is.
    type Compound[A] = T#Compound[H with A]

    type Map[F[_]] = T#Map[F] ∨ F[H]
  }

  def foo[A : (∅ ∨ String ∨ Int ∨ List[Int])#Member](a: A): String = a match {
    case s: String => "String"
    case i: Int => "Int"
    case l: List[_] => "List[Int]"
  }

  foo(42)
  foo("bar")
  foo(List(1, 2, 3))
  foo(42d) // error
  foo[Any](???) // error
}

I did spend some time trying to combine this idea with an upper bound on member types as seen in the TLists of harrah/up, however the implementation of Map with type bounds has thus far proved challenging.