Can I use case object as method parameter - scala

Question

I would like to treat every result differently without using match.
I wanted to make multiple method implementations but it only works for classes, using a case class instead would be too problematic.
The following approach does not work:
sealed trait Result
object Result {
final case object Result1 extends Result
final case object Result2 extends Result
final case object Result3 extends Result
}
def doSth(result: Result)=
consumeResult(result)
private def consumeResult(result: Result.Result1) = ???
private def consumeResult(result: Result.Result2) = ???
private def consumeResult(result: Result.Result3) = ???

Answer 1

The immediate fix is to use the object type like this:
private def consumeResult(result: Result.Result1.type) = ???
etc..
However the broader code is not going to work:
def doSth(result: Result) =
consumeResult(result)
Function overloading happens at compile time, not run time. By the time you call consumeResult the run-time type is lost. The compile-time type is Result so it won't match any of your overloaded versions of consumeResult.
A typeclass might be the best approach:
trait Consume[T <: Result] {
def consume()
}
object Consume {
implicit object consume1 extends Consume[Result.Result1.type] {
def consume() = {
println("Consume1")
}
}
implicit object consume2 extends Consume[Result.Result2.type] {
def consume() = {
println("Consume2")
}
}
}
def doSth[T <: Result](result: T)(implicit ev: Consume[T]) =
ev.consume()
doSth(Result.Result1) // Prints "Consume1"
doSth(Result.Result2) // Prints "Consume2"
doSth(Result.Result3) // Does not compile, no matching typeclass