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

Here is the Rex Kerr way to encode union types. Straight and simple!

scala> def f[A](a: A)(implicit ev: (Int with String) <:< A) = a match {
     |   case i: Int => i + 1
     |   case s: String => s.length
     | }
f: [A](a: A)(implicit ev: <:<[Int with String,A])Int

scala> f(3)
res0: Int = 4

scala> f("hello")
res1: Int = 5

scala> f(9.2)
:9: error: Cannot prove that Int with String <:< Double.
       f(9.2)
        ^

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Source: Comment #27 under this excellent blog post by Miles Sabin which provides another way of encoding union types in Scala.