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.