java - Scala 解析器和组合器 : java. lang.RuntimeException:字符串匹配正则表达式 `\z' 预期

Question

我正在尝试使用 Scala 的 RegexParser 按照动态认知逻辑语法解析一些文本，作为我硕士论文的一部分。但我在简单的逻辑连接上不断遇到同样的错误。我了解它在哪里以及为什么失败，但不知道为什么它与它最初的情况相匹配。

我的代码(严格归结为隔离问题):

import scala.util.parsing.combinator._

class Formula() {
      def and(q:Formula) = Conjunction(this, q) // ∧
}

abstract class Literal extends Formula
abstract class Constant extends Formula

case class Atom(symbol:String) extends Literal 
case class NotAtom(p:Atom) extends Literal

case class Conjunction(p:Formula, q:Formula) extends Formula

class mapParser extends RegexParsers {
    val conjOp = "&"
    val negOp = "~"

    val listseparator = ","

    val leftparen = "("
    val rightparen = ")"

    def id:Parser[String] = "[a-z_]+".r // fluents are never capitalized. but may have underscore.
    def litargs: Parser[String] = repsep("[a-zA-Z]+".r,listseparator) ^^ {case list => "(" + list.toString.stripPrefix("List") + ")"}

    def atom: Parser[Atom] = id~leftparen~litargs~rightparen ^^ {case head~_~tail~_ => Atom(head+tail)}
    def negAtom: Parser[NotAtom] = negOp~>atom ^^ (NotAtom(_))
    def literal: Parser[Literal] = negAtom | atom 

    def and: Parser[Formula] = formula~conjOp~formula ^^ {case p1~_~p2 => Conjunction(p1,p2)}  

    def formula: Parser[Formula] = literal | and
};

object DomainParser extends mapParser {
  def test() =  {
    val domainDesc ="present(A) & ~present(B)";

    println("input: " + domainDesc)
    println("result: " + apply(domainDesc))
  }

  def apply(domainDesc: String) = parseAll(formula, domainDesc) match {
    case Success(result, _) => result
    case failure : NoSuccess => scala.sys.error(failure.msg)
  }
}

我正在从 java 外部调用 DomainParser.test() 函数。输入为

present(A) & ~present(B)

应该产生:

Conjunction(Atom(present((A))),NotAtom(Atom(present((B)))))

但却给了我错误:

Exception in thread "main" java.lang.RuntimeException: string matching regex `\z' expected but `&' found
    at scala.sys.package$.error(package.scala:27)
    at mAp.DomainParser$.apply(DEL.scala:48)
    at mAp.DomainParser$.test(DEL.scala:43)
    at mAp.DomainParser.test(DEL.scala)
at ma.MA.main(MA.java:8)

此外，如果我直接调用“and”解析器而不是“formula”解析器，它可以正常工作。因此问题似乎出在这一行:

def formula: Parser[Formula] = literal | and

因为它尝试将整行解析为单个文字。然后它正确地解析present(A)，但不是在“&”(不是文字解析器的一部分)上失败并返回解析为“and”项，而是失败并出现异常。

我无法理解为什么它会尝试匹配任何“\z”。它没有被我包含在语法中，即使它是 - 它不应该失败并尝试解析为下一个术语而不是异常退出吗？我左右为难，一是认为存在一些我不知道的字符串结尾术语的内置功能，二是认为有一些非常明显的东西就在我面前。

我们非常需要任何帮助，非常欢迎并提前非常感谢您。

丹·特鲁

Answer 1

我只会为我制作的命题公式添加一个类似的解析器。也许这可能对您有帮助。

'+' = 顶部/true

'-' = 底部/假

'!' = 否定

'&' = 连词

'|' = 析取

'>' = 含义

'<' = 等价

object FormulaParser extends StandardTokenParsers with PackratParsers {
  //Symbols for all connectives
  private val parseSymbols = List("(", ")", "+", "-", "!", "&", "|", ">", "<")
  lexical.delimiters ++= parseSymbols

  private lazy val formula: PackratParser[Formula] = implication | equivalence | conjunction | disjunction | term
  private lazy val formulaWithoutBrackets: PackratParser[Formula] = implication | equivalence | conjunction | disjunction | termWithoutBrackets

  private lazy val term: PackratParser[Formula] = top | bottom | variable | parens | negation
  private lazy val termWithoutBrackets = top | bottom | variable | negation

  private lazy val top: PackratParser[Formula] = "+" ^^^ { Top() }
  private lazy val bottom: PackratParser[Formula] = "-" ^^^ { Bottom() }
  private lazy val variable: PackratParser[Formula] = ident ^^ { Variable(_) }
  private lazy val parens: PackratParser[Formula] = "(" ~> formulaWithoutBrackets <~ ")"
  private lazy val negation: PackratParser[Formula] = "!" ~> term ^^ { Negation(_) }

  private lazy val conjunction: PackratParser[Formula] = term ~ "&" ~ term ~ rep("&" ~> term) ^^ {
    case p ~ "&" ~ q ~ conj => conj.foldLeft(Conjunction(p,q))((con, elem) => Conjunction(con, elem))
  }

  private lazy val disjunction: PackratParser[Formula] = term ~ "|" ~ term ~ rep("|" ~> term) ^^ {
    case p ~ "|" ~ q ~ disj => disj.foldLeft(Disjunction(p,q))((dis, elem) => Disjunction(dis, elem))
  }

  private lazy val implication: PackratParser[Formula] = (conjunction | disjunction | term) ~ ">" ~ (conjunction | disjunction | term) ^^ { case p ~ ">" ~ q => Implication(p, q) }

  private lazy val equivalence: PackratParser[Formula] = (conjunction | disjunction | term) ~ "<" ~ (conjunction | disjunction | term) ^^ { case p ~ "<" ~ q => Equivalence(p, q) }
}

有了这个，您可以解析输入，例如: (p & q) | (!q > (r & s))

这里的连词和析取也比蕴涵和等价结合得更强。

p & q > r | s 将导致 Implication(Conjunction(Variable(p), Variable(q)), Disjunction(Variable(r), Variable(s)))