Scala type mismatch - found: Unit required: Array[String]

Question

I do not understand why Scala complains about a type error in the following example:

def GetRanges(RangeString1: String): Array[String] = {
    val GetOneRange = "\\d+\\-\\d+".r;
    var AllRanges = new Array[String](0);
    if (!f_stringNullEmpty(RangeString1)) {
      GetOneRange.findAllIn(RangeString1).matchData.foreach(
        m => AllRanges = AllRanges ++ Array[String](m.group(0)) // Explicit casting to Array[String]
      )
    }
    return scala.util.Sorting.quickSort(AllRanges);
}

the error I receive is:

 notebook:38: error: type mismatch;
 found   : Unit
 required: Array[String]
 return scala.util.Sorting.quickSort(AllRanges);
                                    ^

Apparently, iterating through the regex results and adding them to the array, causes a type change. But why? Or did I miss something more fundamental?

Note: I understand that the if statement returns a type Unit, because no else was specified. But I can;t see that that would affect the type of my array.

Nagarjuna Pamu · Accepted Answer · 2018-07-06 17:23:27Z

1

Return type of scala.util.Sorting.quickSort(AllRanges) is Unit. But, GetRanges requires Array[String]

def GetRanges(RangeString1: String): Array[String] = {
val GetOneRange = "\\d+\\-\\d+".r;

Here is quickSort for sorting Arrays

 /** Sort array `a` with quicksort, using the Ordering on its elements.
    * This algorithm sorts in place, so no additional memory is used aside from
    * what might be required to box individual elements during comparison.
    */
  def quickSort[K: Ordering](a: Array[K]): Unit = {
    // Must have iN >= i0 or math will fail.  Also, i0 >= 0.

answered Jul 6, 2018 at 17:23

Nagarjuna Pamu

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Comments

Joey · Accepted Answer · 2018-07-06 18:11:10Z

0

quickSort() mutates AllRanges in place and just returns Unit, but you've specified that GetRanges() returns Array[String] as if quickSort() was returning an Array (it's not).

You can fix your code by changing it to something like this (nb. you don't need to specify return):

scala.util.Sorting.quickSort(AllRanges)
AllRanges

FWIW, you can also avoid using f_stringNullEmpty and the array concatenation by doing something like this instead:

def getRanges(s: String): Array[String] = {
  val p = """\d+\-\d+""".r
  Option(s).filter(_.nonEmpty).map(p.findAllIn) match {
    case Some(matches) if matches.nonEmpty =>
      val m = matches.toArray[String]
      scala.util.Sorting.quickSort(m)
      m
    case _ =>
      Array.empty[String]
  }
}

answered Jul 6, 2018 at 18:11

Joey

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1 Comment

M.S.Visser Over a year ago

Joey, thanks. I expected quickSort to return the ordered array, which indeed it does not. FYI The position of ^ in the error message got me to think that the argument to quickSort was of the wrong type, not that the return type of quickSort was of the wrong type. Learning day by day...

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