forall in Scala

Question

As shown below, in Haskell, it\'s possible to store in a list values with heterogeneous types with certain context bounds on them:

data ShowBox = forall s. S         


        

Answer 1

I don't think a 1-to-1 translation from Haskell to Scala is possible here. But why don't you want to use subtyping? If the types you want to use (such as Int) lack a show method, you can still add this via implicit conversions.

scala> trait Showable { def show:String }
defined trait Showable

scala> implicit def showableInt(i:Int) = new Showable{ def show = i.toString }
showableInt: (i: Int)java.lang.Object with Showable

scala> val l:List[Showable] = 1::Nil
l: List[Showable] = List($anon$1@179c0a7)

scala> l.map(_.show)
res0: List[String] = List(1)