Question

EDIT: This was an old bug long since fixed in Scala 2.8 and later

During some experimentation around question Pattern matching a String as Seq[Char], I ran across another weird matching phenomenon. Consider the following code that treats a string as a sequence of characters:

def %%&#(input: String) : String =  {     val uha : Seq[Char] = input     uha match {         case Seq() => 'Empty'         case Seq(first @ _, 'o', 'o')  => 'Bar'         case _ => 'Oh'      } } 

Calling input on the empty String '' correctly yields 'Empty'.

However, if I rewrite the first match clause as

case Seq.empty => 'Empty' 

the matching of '' fails and matches the default clause instead.

Walking through the Scala library source code (which you shouldn’t have to do in an ideal world 🙂 ) I believe that both Seq() and Seq.empty will result in RandomAccessSeq.empty. Apparently, this does not concur with the phenomenon described above because only Seq() matches the empty String.

UPDATE: Upon some further experimentation this question can be narrowed down to the following:

val list = List()    >>> list2: List[Nothing] = List() val emptySeq = Seq.empty list == emptySeq    >>> res1: Boolean = false 

This basically means that an empty Seq does not automatically equal Seq.empty . So when matching against a constant (as opposed to using an extractor as suggested by starblue) this unequality leads to the failing match. The same is true when interpreting the empty String as a sequence.

Answer 1

This appears to be a bug in the library. Do you want to file the bug or shall I?

scala> Seq.empty  match {case Seq() => 'yup'; case _ => 'nope'} res0: java.lang.String = yup  scala> Seq()  match {case Seq.empty => 'yup'; case _ => 'nope'} res1: java.lang.String = yup  scala> ('' : Seq[Char]) match {case Seq() => 'yup'; case _ => 'nope'}     res2: java.lang.String = yup  scala> ('' : Seq[Char]) match {case Seq.empty => 'yup'; case _ => 'nope'} res3: java.lang.String = nope