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?

flavian · Accepted Answer · 2016-06-05 02:19:31Z

3

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.

answered Jun 5, 2016 at 2:19

flavian

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