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)
}
*
,+
, and?
operators always behave greedily, consuming as much input as possible and never backtracking: Expressiona*
will always consume as many a's as are consecutively available in the input string, causing(a* a)
to fail always. – Hone