解析“elem”方法前面的空白

Question

我试图解析两个Int和一些元素的输入，最后：

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")
      }
    }
}

方法go中给出了一个示例输入。Ints和末尾的元素必须用空格分隔。但这只适用于Ints，而不适用于元素。当我输入时

val x = new X
x go x.p1

我得到以下错误：

java.lang.RuntimeException: string matching regex `z' expected but `W' found

但是当我输入时

x go x.p1

我得到：

java.lang.RuntimeException: string matching regex `s+' expected but `W' found

最后我想要一个Parser[Int ~ Int ~ List[List[Char]]]。为什么在elem前面插入空格无效？我该如何让这些代码发挥作用？

Answer 1

只需用RegEx Parser:替换elems

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")
  }
}
}

我已经删除了p2，因为现在没有用处