I wrote a new combiner for my parser in scala.
This is a ^^ combinator variation that conveys location information. But access to information about the position of the input element is really costly.
In my case, parsing a large example takes about 3 seconds without location information, since it takes more than 30 seconds.
I wrote a runnable example where the execution time is about 50% longer when accessing a position.
Why? How can I get the best lead time?
Example:
import scala.util.parsing.combinator.RegexParsers
import scala.util.parsing.combinator.Parsers
import scala.util.matching.Regex
import scala.language.implicitConversions
object FooParser extends RegexParsers with Parsers {
var withPosInfo = false
def b: Parser[String] = regexB("""[a-z]+""".r) ^^@ { case (b, x) => b + " ::" + x.toString }
def regexB(p: Regex): BParser[String] = new BParser(regex(p))
class BParser[T](p: Parser[T]) {
def ^^@[U](f: ((Int, Int), T) => U): Parser[U] = Parser { in =>
val source = in.source
val offset = in.offset
val start = handleWhiteSpace(source, offset)
val inwo = in.drop(start - offset)
p(inwo) match {
case Success(t, in1) =>
{
var a = 3
var b = 4
if(withPosInfo)
{ // takes a lot of time
a = inwo.pos.line
b = inwo.pos.column
}
Success(f((a, b), t), in1)
}
case ns: NoSuccess => ns
}
}
}
def main(args: Array[String]) = {
val r = "foo"*50000000
var now = System.nanoTime
parseAll(b, r)
var us = (System.nanoTime - now) / 1000
println("without: %d us".format(us))
withPosInfo = true
now = System.nanoTime
parseAll(b, r)
us = (System.nanoTime - now) / 1000
println("with : %d us".format(us))
}
}
Conclusion:
without: 2952496 us
s: 4591070 us
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