I wrote the following Parser with the intent of fail
-ing on whitespace:
import scala.util.parsing.combinator._
object Foo extends JavaTokenParsers {
val wsTest = not(whiteSpace) // uses whitespace inherited from `RegexParsers`
}
Why is parsing a bunch of whitespace successfull?
scala> Foo.parseAll(Foo.wsTest, " ")
res5: Foo.ParseResult[Unit] = [1.11] parsed: ()
scala> res5.successful
res6: Boolean = true
Looking at Parsers#not
from the project , I would've expected a Failure
for my above test.
/** Wrap a parser so that its failures and errors become success and
* vice versa -- it never consumes any input.
*/
def not[T](p: => Parser[T]): Parser[Unit] = Parser { in =>
p(in) match {
case Success(_, _) => Failure("Expected failure", in)
case _ => Success((), in)
}
}
JavaTokenParsers extends RegexParsers, RegexParsers has:
protected val whiteSpace = """\s+""".r
def skipWhitespace = whiteSpace.toString.length > 0
implicit def regex(r: Regex): Parser[String] = new Parser[String] {
...
val start = handleWhiteSpace(source, offset)
...
}
protected def handleWhiteSpace(source: java.lang.CharSequence, offset: Int): Int =
if (skipWhitespace)
(whiteSpace findPrefixMatchOf (source.subSequence(offset, source.length))) match {
case Some(matched) => offset + matched.end
case None => offset
}
else
offset
so it skips whitespace (you can override this by overriding def skipWhitespace = false)
so for the parser " "
equals ""
whitespace tries to match "" but it fails ("""\\s+""" requires at least one whitespace) and the not converts this in a success
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