Default value for type parameter in Scala

Question

I'm failing to figure out how (if at all) you can set a default value for a type -parameter in Scala.
Currently I have a method similar to this:

def getStage[T <: Stage](key: String): T = {
  // Do fancy stuff that returns something
}

But what I'd like to do is provide an implementation of getStage that takes no value for T and uses a default value instead. I tried to just define another method and overload the parameters, but it only leads to one of the methods being completely overriden by the other one. If I have not been clear what I'm trying to do is something like this:

def getStage[T<:Stage = Stage[_]](key: String): T = {

}

I hope it's clear what I'm asking for. Does anyone know how something like this could be achieved?

Answer 1

You can do this kind of thing in a type-safe way using type classes. For example, suppose you've got this type class:

trait Default[A] { def apply(): A }

And the following type hierarchy:

trait Stage
case class FooStage(foo: String) extends Stage
case class BarStage(bar: Int) extends Stage

And some instances:

trait LowPriorityStageInstances {
  implicit object barStageDefault extends Default[BarStage] {
    def apply() = BarStage(13)
  }
}

object Stage extends LowPriorityStageInstances {
  implicit object stageDefault extends Default[Stage] {
    def apply() = FooStage("foo")
  }
}

Then you can write your method like this:

def getStage[T <: Stage: Default](key: String): T =
  implicitly[Default[T]].apply()

And it works like this:

scala> getStage("")
res0: Stage = FooStage(foo)

scala> getStage[BarStage]("")
res1: BarStage = BarStage(13)

Which I think is more or less what you want.