Question

def test1(a: Any) = a match {
  case x: AnyRef => "AnyRef"
  case _ => "None of the above"
}

def test2(a: Any) = a match {
  case x: Double if x > 2 => "Double > 2"
  case x: AnyRef => "AnyRef"
  case _ => "None of the above"
}

Please can someone explain why in the following, the first case 1.0 matches on AnyRef, but in the second it doesn’t. (Scala 2.9.0-1)

scala> test1(1.0)
res28: java.lang.String = AnyRef

scala> test2(1.0)
res29: java.lang.String = None of the above

---

edit – Scala 2.10 update Jan 2013: the new pattern matcher fixes this behaviour (or at least, makes it consistent) and the method test2 now returns “AnyRef” as for test1.

Answer 1

This is because Any is actually just an Object. Having Double there is a convenient fiction–it’s actually java.lang.Double which is autounboxed for you in the match statement. Unfortunately, there is no way for Scala to tell if it finds a java.lang.Double if it is supposed to be interpreted as a Double or as a java.lang.Double–in the latter case, the AnyRef should catch it. So it does. But if you specifically ask for a Double, it knows it is supposed to unbox, and then the AnyRef case need not be checked. (And, in fact, if you intended it to be a java.lang.Double, it will unbox that too–it can’t tell the difference.)

Whether this is ideal behavior is debatable, but it is logical.