non-greedy matching in Scala RegexParsers

问题

Suppose I'm writing a rudimentary SQL parser in Scala. I have the following:

class Arith extends RegexParsers {
    def selectstatement: Parser[Any] = selectclause ~ fromclause
    def selectclause: Parser[Any] = "(?i)SELECT".r ~ tokens
    def fromclause: Parser[Any] = "(?i)FROM".r ~ tokens
    def tokens: Parser[Any] = rep(token) //how to make this non-greedy?
    def token: Parser[Any] = "(\\s*)\\w+(\\s*)".r
}

When trying to match selectstatement against SELECT foo FROM bar, how do I prevent the selectclause from gobbling up the entire phrase due to the rep(token) in ~ tokens?

In other words, how do I specify non-greedy matching in Scala?

To clarify, I'm fully aware that I can use standard non-greedy syntax (*?) or (+?) within the String pattern itself, but I wondered if there's a way to specify it at a higher level inside def tokens. For example, if I had defined token like this:

def token: Parser[Any] = stringliteral | numericliteral | columnname

Then how can I specify non-greedy matching for the rep(token) inside def tokens?

回答1:

Not easily, because a successful match is not retried. Consider, for example:

object X extends RegexParsers {
  def p = ("a" | "aa" | "aaa" | "aaaa") ~ "ab"
}

scala> X.parseAll(X.p, "aaaab")
res1: X.ParseResult[X.~[String,String]] = 
[1.2] failure: `ab' expected but `a' found

aaaab
 ^

The first match was successful, in parser inside parenthesis, so it proceeded to the next one. That one failed, so p failed. If p was part of alternative matches, the alternative would be tried, so the trick is to produce something that can handle that sort of thing.

Let's say we have this:

def nonGreedy[T](rep: => Parser[T], terminal: => Parser[T]) = Parser { in =>
  def recurse(in: Input, elems: List[T]): ParseResult[List[T] ~ T] =
    terminal(in) match {
      case Success(x, rest) => Success(new ~(elems.reverse, x), rest)
      case _ => 
        rep(in) match {
          case Success(x, rest) => recurse(rest, x :: elems)
          case ns: NoSuccess    => ns
        }
    }

  recurse(in, Nil)
}

You can then use it like this:

def p = nonGreedy("a", "ab")

By the way,I always found that looking at how other things are defined is helpful in trying to come up with stuff like nonGreedy above. In particular, look at how rep1 is defined, and how it was changed to avoid re-evaluating its repetition parameter -- the same thing would probably be useful on nonGreedy.

Here's a full solution, with a little change to avoid consuming the "terminal".

trait NonGreedy extends Parsers {
    def nonGreedy[T, U](rep: => Parser[T], terminal: => Parser[U]) = Parser { in =>
      def recurse(in: Input, elems: List[T]): ParseResult[List[T]] =
        terminal(in) match {
          case _: Success[_] => Success(elems.reverse, in)
          case _ => 
            rep(in) match {
              case Success(x, rest) => recurse(rest, x :: elems)
              case ns: NoSuccess    => ns
            }
        }

      recurse(in, Nil)
    }  
}

class Arith extends RegexParsers with NonGreedy {
    // Just to avoid recompiling the pattern each time
    val select: Parser[String] = "(?i)SELECT".r
    val from: Parser[String] = "(?i)FROM".r
    val token: Parser[String] = "(\\s*)\\w+(\\s*)".r
    val eof: Parser[String] = """\z""".r

    def selectstatement: Parser[Any] = selectclause(from) ~ fromclause(eof)
    def selectclause(terminal: Parser[Any]): Parser[Any] = 
      select ~ tokens(terminal)
    def fromclause(terminal: Parser[Any]): Parser[Any] = 
      from ~ tokens(terminal)
    def tokens(terminal: Parser[Any]): Parser[Any] = 
      nonGreedy(token, terminal)
}

来源：https://stackoverflow.com/questions/7812610/non-greedy-matching-in-scala-regexparsers

标签

regex

scala

parser-generator

non-greedy