parsing white spaces in front of `elem` method - scala

Question

I tried to parse an input of two Ints and some elements and the end:
import scala.util.parsing.combinator.JavaTokenParsers
class X extends JavaTokenParsers {
lazy val elems = elem("wrong elem", "#WB-" contains _)
lazy val lists = repsep(rep(elems), ",")
lazy val p1 = int ~ int ~ lists
lazy val p2 = int ~ int ~ (whiteSpace ~> lists)
def go[A](p: Parser[A]) = parseAll(p, "1 2 WB#,---,BBB") match {
case NoSuccess(msg, _) => sys.error(msg)
case _ =>
}
lazy val int: Parser[Int] =
wholeNumber ^^ {
try _.toInt catch {
case e: NumberFormatException => sys.error("invalid number")
}
}
}
An example input is given in method go. The Ints and the elements at the end have to be delimited by spaces. But this works only for the Ints and not for the elements. When I type in
val x = new X
x go x.p1
I get following error:
java.lang.RuntimeException: string matching regex `\z' expected but `W' found
But when I type in
x go x.p1
I get:
java.lang.RuntimeException: string matching regex `\s+' expected but `W' found
At the end I want to have a Parser[Int ~ Int ~ List[List[Char]]]. Why does inserting white spaces in front of elem not work? And how can I get this code to work?

Answer 1

Just replace elems by a RegEx Parser :
import scala.util.parsing.combinator.JavaTokenParsers
class X extends JavaTokenParsers {
lazy val elems = "[#WB-]".r
lazy val lists = repsep(rep(elems), ",")
lazy val p1 = int ~ int ~ lists
def go[A](p: Parser[A]) = parseAll(p, "1 2 WB#,---,BBB") match {
case NoSuccess(msg, _) => sys.error(msg)
case _ =>
}
lazy val int: Parser[Int] =
wholeNumber ^^ {
try _.toInt catch {
case e: NumberFormatException => sys.error("invalid number")
}
}
}
i have removed p2 because is not useful now