scala - How do I match types that don't have a specific typeclass instance? -
this question has answer here:
i define behavior types don't have instance typeclass:
// given trait sometypeclass[t] // when have implicit sometypeclass[t] def f[t: sometypeclass](x:t):unit = ??? // when don't have instance def f[t !: sometypeclass](x: t):unit = ??? we handle difference within typeclass need create instances support generic behavior.
is there way negate type bound? way make function !: compile?
(i in vanilla scala, without scalaz, shapeless, etc)
is there way negate type bound?
no! syntax [t: sometypeclass] shorthand (implicit val t: sometypeclass[t]), , there no way "negate" that. overload method create ambiguity.
but can "nest" type classes.
trait existslowpri { implicit def no[a]: exists[a] = exists.no } object exists extends existslowpri { case class yes[a](peer: a) extends exists[a] case object no extends exists[nothing] implicit def yes[a](implicit peer: a): exists[a] = new yes(peer) } sealed trait exists[+a] example:
trait show[-a] { def show(x: a): string } def test[a](x: a)(implicit ex: exists[show[a]]): unit = println( ex match { case exists.yes(s) => s.show(x) case exists.no => "(no string repr)" } ) implicit object showboolean extends show[boolean] { def show(b: boolean) = if (b) "t" else "f" } test(123) // (no string repr) test(true) // t however, advise against doing this, because main point of implicits , type classes have explicit compiler failure if not in scope. way able compile have no guarantee correctly brought specific type class scope.
