Why can't I pattern match on Stream.empty in Scala?

Question

The code below doesn't compile if I uncomment the line indicated. The compiler complains: "stable identifier required".

val Empty = Stream.empty    
val a = Stream.range(0, 5)
a match {
  // case Stream.empty => println("nope") <-- does not work
  case Empty => println("compiles") <-- works
  case _ => println("ok")
}

If I assign Stream.empty to value Empty first, it works, but it feels strange that you can't pattern match on such a fundamental value without such a hack.

Am I missing something?

Marth · Accepted Answer · 2017-05-06T17:56:33.977

50

You can't pattern match on Stream.empty because it is a method (in object Stream) that always returns the empty stream (but the compiler doesn't know that).

Instead of assigning val empty = Stream.empty, you can match on Stream.Empty, which is an Object :

scala> a match {
           case Stream.Empty => println("done")
           case h #:: tl => println(h)
       }

edited May 06 '17 at 17:56

answered Feb 15 '15 at 20:49

Marth

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Confirmed it works for Scala 2.12, at least. Things may be different starting from 2.13. – ragazzojp Mar 29 '20 at 16:39

score 6 · Answer 2 · edited May 23 '17 at 11:54

You can't "pattern match" on a variable that is not a constant.
Stream.empty is not a "stable" identifier since it represents some method:

/** The empty stream */
  override def empty[A]: Stream[A] = Empty

that could potentially return any value at any time.
Compiler isn't aware that its returned value is always Empty, so it detects it as a potential changing variable.
Too deep for it to detect it.

However, when you assign the method's retult to a val (being a stable identifier since immutable), your code is able to process to pattern matching using it.

You might read this, evoking an hypothesis explaining why pattern matching expects a stable identifier.

Why can't I pattern match on Stream.empty in Scala?

2 Answers2

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