CODEWITHSUNDEEP
xmlscala
Kevin Meredith
Kevin Meredith

Reputation: 41939

Pattern Matching on scala.xml.NodeSeq

Given:

$cat build.sbt
scalaVersion := "2.11.8"

libraryDependencies += "org.scala-lang.modules" %% "scala-xml"   % "1.0.6"

and, then 

$sbt console

def exactlyOne[A](xs: Seq[A]): Option[A] = xs match {
 case head :: Nil => Some(head)
 case _           => None
}

scala> val xml = <root><a/></root>
xml: scala.xml.Elem = <root><a/></root>

scala> xml \ "a"
res3: scala.xml.NodeSeq = NodeSeq(<a/>)

scala> exactlyOne( res3 )
res4: Option[scala.xml.Node] = None

Evidently, Seq#unapply is not being used:

scala> exactlyOne( Seq(1) )
res2: Option[Int] = Some(1)

My understanding is that unapply will get called, typically on the class's companion object.

I found Node#unapplySeq (http://www.scala-lang.org/api/2.11.8/scala-xml/index.html#scala.xml.Node$@unapplySeq(n:scala.xml.Node):Some[(String,scala.xml.MetaData,Seq[scala.xml.Node])]), but I'm not sure if that's being called.

Which method is getting called on match here?

Upvotes: 0

Views: 410

Answers (1)

Michael Zajac
Michael Zajac

Reputation: 55569

A NodeSeq is not a List, so trying to pattern match it as one will fail. If you wanted to match it using unapplySeq of some kind, you would need to do it in this way:

def exactlyOne[A](xs: Seq[A]): Option[A] = xs match {
  case Seq(head) => Some(head)
  case _ => None
}

scala> exactlyOne(<root>hello</root>)
res5: Option[scala.xml.Node] = Some(<root>hello</root>)

scala> exactlyOne(<root>hello</root><foo>world</foo>)
res6: Option[scala.xml.Node] = None

Upvotes: 3

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