用于递归 bnf 的 Scala Parser Combinators 技巧？

Question

我试图匹配这个语法:

pgm ::= exprs
exprs ::= expr [; exprs]
expr ::= ID | expr . [0-9]+

我的 Scala Packrat 解析器组合器如下所示:

import scala.util.parsing.combinator.PackratParsers
import scala.util.parsing.combinator.syntactical._

object Dotter extends StandardTokenParsers with PackratParsers {
    lexical.delimiters ++= List(".",";")
    def pgm = repsep(expr,";")
    def expr :Parser[Any]= ident | expr~"."~num
    def num = numericLit

       def parse(input: String) =
    phrase(pgm)(new PackratReader(new lexical.Scanner(input))) match {
      case Success(result, _) => println("Success!"); Some(result)
      case n @ _ => println(n);println("bla"); None
    }  

    def main(args: Array[String]) {
      val prg = "x.1.2.3;" +
            "y.4.1.1;" +
            "z;" +
            "n.1.10.30"


            parse(prg);
    }
}

但这不起作用。要么“匹配贪婪”并告诉我:

[1.2] failure: end of input expected 
x.1.2.3;y.4.1.1;z;n.1.10.30

或者如果我更改 |到 |||我得到一个堆栈溢出:

Exception in thread "main" java.lang.StackOverflowError
at java.lang.Character.isLetter(Unknown Source)
at java.lang.Character.isLetter(Unknown Source)
at scala.util.parsing.combinator.lexical.Lexical$$anonfun$letter$1.apply(Lexical.scala:32)
at scala.util.parsing.combinator.lexical.Lexical$$anonfun$letter$1.apply(Lexical.scala:32)
...

我有点明白为什么我会收到错误；我该怎么做才能解析像上面这样的语法？对我来说似乎并不深奥

编辑:
基于 http://scala-programming-language.1934581.n4.nabble.com/Packrat-parser-guidance-td1956908.html 中引用的论文
我发现我的程序实际上并没有使用新的 Packrat 解析器。

IE。改 Parser[Any]至 PackratParser[Any]并使用 lazy val而不是 def
我将上面的内容改写为:

import scala.util.parsing.combinator.PackratParsers
import scala.util.parsing.combinator.syntactical._

object Dotter extends StandardTokenParsers with PackratParsers {
    lexical.delimiters ++= List(".",";")
    lazy val pgm : PackratParser[Any] = repsep(expr,";")
    lazy val expr :PackratParser[Any]= expr~"."~num | ident
    lazy val num = numericLit

    def parse(input: String) =
    phrase(pgm)(new PackratReader(new lexical.Scanner(input))) match {
      case Success(result, _) => println("Success!"); Some(result)
      case n @ _ => println(n);println("bla"); None
    }  

    def main(args: Array[String]) {
      val prg = "x.1.2.3 ;" +
            "y.4.1.1;" +
            "z;" +
            "n.1.10.30"


            parse(prg);
    }
}

Answer 1

问题是(至少部分)您实际上并没有使用 Packrat 解析器。请参阅 Scala 的文档 PackratParsers特质，它说

Using PackratParsers is very similar to using Parsers:

• any class/trait that extends Parsers (directly or through a subclass) can mix in PackratParsers. Example: object MyGrammar extends StandardTokenParsers with PackratParsers
• each grammar production previously declared as a def without formal parameters becomes a lazy val, and its type is changed from Parser[Elem] to PackratParser[Elem]. So, for example, def production: Parser[Int] = {...} becomes lazy val production: PackratParser[Int] = {...}
• Important: using PackratParsers is not an all or nothing decision. They can be free mixed with regular Parsers in a single grammar.

我对 Scala 2.8 的解析器组合器了解不够，无法完全解决这个问题，但是通过以下修改，我能够解析到分号，这是对您所做的改进的改进。

object Dotter extends StandardTokenParsers with PackratParsers {
    lexical.delimiters ++= List(".",";")
    lazy val pgm:PackratParser[Any] = repsep(expr,";")
    lazy val expr:PackratParser[Any]= ident ||| (expr~"."~numericLit)

    def parse(input: String) = phrase(expr)(lex(input)) match {
      case Success(result, _) => println("Success!"); Some(result)
      case n @ _ => println(n);println("bla"); None
    }  

    def lex(input:String) = new PackratReader(new lexical.Scanner(input))
}