Ok, I know better than to use nulls as a design choice, but in this case I have to. Why the following does not compile?
def test[T<:AnyRef](o :Option[T]) :T = o getOrElse null Error:(19, 53) type mismatch; found : Null(null) required: T Note: implicit method foreignKeyType is not applicable here because it comes after the application point and it lacks an explicit result type def test[T<:AnyRef](o :Option[T]) :T = o getOrElse null ^ Null is a subtype of all reference types, but the fact that T is a subtype of AnyRef doesn't guarantee that T is a reference type -- in particular, Nothing is a subtype of AnyRef which does not contain null.
Your code Works if you add a lower bound:
def test[T >:Null <:AnyRef](o :Option[T]) :T = o getOrElse null; It works:
scala> def test[T >:Null <:AnyRef](o :Option[T]) :T = o getOrElse null; test: [T >: Null <: AnyRef](o: Option[T])T scala> scala> scala> test(None) res0: Null = null scala> test(Some(Some)) res1: Some.type = Some 
Nothing <: Null. Otherwise, you could create an instance of Nothing using test[Nothing](None).I don't know why this does not work - Null is a subtype of all reference types in Scala, so you would expect that this would work with any T <: AnyRef. You can make it work with asInstanceOf:
def test[T <: AnyRef](o: Option[T]): T = o getOrElse null.asInstanceOf[T] (Try to avoid using null whenever possible in Scala - I can imagine you'd have a legitimate use case, for example when you need to pass data to Java code).
By the way, Option has a method orNull which will return the value of an option if it is a Some and null if it is None.
test(None: Option[String])
Option.orNull?