Scala:我可以将组合器解析器微调为本地贪婪吗? [英] Scala: Can I nudge a combinator parser to be locally greedy?

问题描述

假设我在组合器解析器中表达的语言不明确.有没有办法使某些表达在局部上是贪婪的?这是我的意思的例子.

Suppose I have an ambiguous language expressed in combinator parser. Is there a way to make certain expressions locally greedy? Here's an example of what I mean.

import scala.util.parsing.combinator._

object Example extends JavaTokenParsers {
  def obj: Parser[Any] = (shortchain | longchain) ~ anyrep

  def longchain: Parser[Any] = zero~zero~one~one
  def shortchain: Parser[Any] = zero~zero

  def anyrep: Parser[Any] = rep(any)
  def any: Parser[Any] = zero | one
  def zero: Parser[Any] = "0"
  def one: Parser[Any] = "1"
  def main(args: Array[String]) {
    println(parseAll(obj, args(0) ))
  }
}

编译后，可以如下运行:

After compiling, I can run it as follows:

$ scala Example 001111
[1.7] parsed: ((0~0)~List(1, 1, 1, 1))

我想以某种方式指示<​​c0>的第一部分是局部贪婪并与longchain匹配.如果我改变顺序，它会匹配longchain，但这不是因为贪婪.

I would like to somehow instruct the first part of obj to be locally greedy and match with longchain. If I switch the order around, it matches the longchain, but that's not because of the greediness.

def obj: Parser[Any] = (longchain | shortchain) ~ anyrep

推荐答案

使用|||:

object Example extends JavaTokenParsers {
  def obj: Parser[Any] = (shortchain ||| longchain) ~ anyrep

  def longchain: Parser[Any] = zero~zero~one~one
  def shortchain: Parser[Any] = zero~zero

  def anyrep: Parser[Any] = rep(any)
  def any: Parser[Any] = zero | one
  def zero: Parser[Any] = "0"
  def one: Parser[Any] = "1"
  def main(args: Array[String]) {
    println(parseAll(obj, args(0) ))
  }
}

scala> Example.main(Array("001111"))
[1.7] parsed: ((((0~0)~1)~1)~List(1, 1))

这篇关于Scala:我可以将组合器解析器微调为本地贪婪吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！