Why Scala fail to recognize Null type as a subtype of T in generic function type parameter inference?

Question

I have the following function definition in scala:

trait GenExtractor[+R] P
      def orElseExtractor[R2<: Super, Super>: R](g: GenExtractor[T, R2]): GenExtractor[T, Super] =
        new OrElse[T, Super](this, g)
}

which should combine 2 GenExtractors:

GenExtractor[A]
GenExtractor[B]

into:

GenExtractor[C]

where C is the common supertype of A and B

However, when I try to invoke this function:

val genExtractor = new GenExtractor[R](...parameters...)
val combined = genExtractor.orElseExtractor[Null, R] {
      _: FetchedRow => null
    }

I got the following error:

Error:(84, 47) type arguments [Null,R] do not conform to method orElseExtractor's type parameter bounds [R2 <: Super,Super >: R]
    def orNull: Extractor[R] = orElseExtractor[Null, R] {
                                              ^

This is clearly a false alarm since in this case:

type R2 = Null
type Super = R

which fulfil the condition: Null <: R & R >: R

Why is scala compiler gave me this error? What should I do to fix it?

Answer 1

Why would Null be a subtype of a generic R ? It's not the compiler being off the mark, it's the underlying assumption that Null <: R is always true. Here are a few fun examples, and they have to do with primitives.

Some(5).orNull will yield error: Cannot prove that Null <:< Int

Here's your gap, you can try it out in the REPL:

implicitly[Null <:< AnyRef] // will compile
implicitly[Null <:< AnyVal] // blows up, primitives are not included.

In the type system, the supertype of all types, the only thing guaranteed to satisfy any such relationship as you expect it, is scala.Nothing , not scala.Null .

null on the JVM is designed mostly to cope with the absence of type information, not to deal with type system hierarchy. For that you have Nothing , AnyRef , AnyVal and all the other fun stuff.