How to define "type disjunction" (union types)?

Well, in the specific case of Any*, this trick below won't work, as it will not accept mixed types. However, since mixed types wouldn't work with overloading either, this may be what you want.

First, declare a class with the types you wish to accept as below:

class StringOrInt[T]
object StringOrInt {
  implicit object IntWitness extends StringOrInt[Int]
  implicit object StringWitness extends StringOrInt[String]
}

Next, declare foo like this:

object Bar {
  def foo[T: StringOrInt](x: T) = x match {
    case _: String => println("str")
    case _: Int => println("int")
  }
}

And that's it. You can call foo(5) or foo("abc"), and it will work, but try foo(true) and it will fail. This could be side-stepped by the client code by creating a StringOrInt[Boolean], unless, as noted by Randall below, you make StringOrInt a sealed class.

It works because T: StringOrInt means there's an implicit parameter of type StringOrInt[T], and because Scala looks inside companion objects of a type to see if there are implicits there to make code asking for that type work.

Miles Sabin describes a very nice way to get union type in his recent blog post Unboxed union types in Scala via the Curry-Howard isomorphism:

He first defines negation of types as

type ¬[A] = A => Nothing

using De Morgan's law this allows him to define union types

type ∨[T, U] = ¬[¬[T] with ¬[U]]

With the following auxiliary constructs

type ¬¬[A] = ¬[¬[A]]
type |∨|[T, U] = { type λ[X] = ¬¬[X] <:< (T ∨ U) }

you can write union types as follows:

def size[T : (Int |∨| String)#λ](t : T) = t match {
    case i : Int => i
    case s : String => s.length
}