scala – 如何忽略解析器组合器中的单行注释

val p = "(" ~> "[a-z]*".r <~ ")"

这将解析类似(abc)的东西,但由于评论你可能会遇到类似的东西：

( ; comment goes here
  abc
)

然后我建议使用TokenParser或其子类之一.这是更多的工作,因为你必须提供一个词法解析器,它将执行第一次传递以丢弃注释.但是如果你有嵌套的评论或者如果有的话,它也会更灵活;可以逃脱或如果;可以在字符串文字内,如：

abc = "; don't ignore this" ; ignore this

另一方面,您也可以尝试覆盖whitespace的值

override protected val whiteSpace = """(s|;.*)+""".r

或类似的规定.
例如,使用RegexParsers scaladoc中的示例：

import scala.util.parsing.combinator.RegexParsers

object so1 {
  Calculator("""(1 + ; foo
  (1 + 2))
  ; bar""")
}

object Calculator extends RegexParsers {
  override protected val whiteSpace = """(s|;.*)+""".r
  def number: Parser[Double] = """d+(.d*)?""".r ^^ { _.toDouble }
  def factor: Parser[Double] = number | "(" ~> expr <~ ")"
  def term: Parser[Double] = factor ~ rep("*" ~ factor | "/" ~ factor) ^^ {
    case number ~ list => (number /: list) {
      case (x,"*" ~ y) => x * y
      case (x,"/" ~ y) => x / y
    }
  }
  def expr: Parser[Double] = term ~ rep("+" ~ log(term)("Plus term") | "-" ~ log(term)("Minus term")) ^^ {
    case number ~ list => list.foldLeft(number) { // same as before,using alternate name for /:
      case (x,"+" ~ y) => x + y
      case (x,"-" ~ y) => x - y
    }
  }
  def apply(input: String): Double = parseAll(expr,input) match {
    case Success(result,_) => result
    case failure: NoSuccess => scala.sys.error(failure.msg)
  }
}

这打印：

Plus term --> [2.9] parsed: 2.0
Plus term --> [2.10] parsed: 3.0
res0: Double = 4.0