scala parser combinator infinite loop - scala

Question

I'm trying to write a simple parser in scala but when I add a repeated token Scala seems to get stuck in an infinite loop.
I have 2 parse methods below. One uses rep(). The non repetitive version works as expected (not what I want though) using the rep() version results in an infinite loop.
EDIT:
This was a learning example where I tired to enforce the '=' was surrounded by whitespace.
If it is helpful this is my actual test file:
a = 1
b = 2
c = 1 2 3
I was able to parse: (with the parse1 method)
K = V
but then ran into this problem when tried to expand the exercise out to:
K = V1 V2 V3
import scala.util.parsing.combinator._
import scala.io.Source.fromFile
class MyParser extends RegexParsers {
override def skipWhitespace(): Boolean = { false }
def key: Parser[String] = """[a-zA-Z]+""".r ^^ { _.toString }
def eq: Parser[String] = """\s+=\s+""".r ^^ { _.toString.trim }
def string: Parser[String] = """[^ \t\n]*""".r ^^ { _.toString.trim }
def value: Parser[List[String]] = rep(string)
def foo(key: String, value: String): Boolean = {
println(key + " = " + value)
true
}
def parse1: Parser[Boolean] = key ~ eq ~ string ^^ { case k ~ eq ~ string => foo(k, string) }
def parse2: Parser[Boolean] = key ~ eq ~ value ^^ { case k ~ eq ~ value => foo(k, value.toString) }
def parseLine(line: String): Boolean = {
parse(parse2, line) match {
case Success(matched, _) => true
case Failure(msg, _) => false
case Error(msg, _) => false
}
}
}
object TestParser {
def usage() = {
System.out.println("<file>")
}
def main(args: Array[String]) : Unit = {
if (args.length != 1) {
usage()
} else {
val mp = new MyParser()
fromFile(args(0)).getLines().foreach { mp.parseLine }
println("done")
}
}
}

Answer 1

Next time, please provide some concrete examples, it's not obvious what your input is supposed to look like.
Meanwhile, you can try this, maybe you find it helpful:
import scala.util.parsing.combinator._
import scala.io.Source.fromFile
class MyParser extends JavaTokenParsers {
// override def skipWhitespace(): Boolean = { false }
def key: Parser[String] = """[a-zA-Z]+""".r ^^ { _.toString }
def eq: Parser[String] = "="
def string: Parser[String] = """[^ \t\n]+""".r
def value: Parser[List[String]] = rep(string)
def foo(key: String, value: String): Boolean = {
println(key + " = " + value)
true
}
def parse1: Parser[Boolean] = key ~ eq ~ string ^^ { case k ~ eq ~ string => foo(k, string) }
def parse2: Parser[Boolean] = key ~ eq ~ value ^^ { case k ~ eq ~ value => foo(k, value.toString) }
def parseLine(line: String): Boolean = {
parseAll(parse2, line) match {
case Success(matched, _) => true
case Failure(msg, _) => false
case Error(msg, _) => false
}
}
}
val mp = new MyParser()
for (line <- List("hey = hou", "hello = world ppl", "foo = bar baz blup")) {
println(mp.parseLine(line))
}
Explanation:
JavaTokenParsers and RegexParsers treat white space differently.
The JavaTokenParsers handles the white space for you, it's not specific for Java, it works for most non-esoteric languages. As long as you are not trying to parse Whitespace, JavaTokenParsers is a good starting point.
Your string definition included a *, which caused the infinite recursion.
Your eq definition included something that messed with the empty space handling (don't do this unless it's really necessary).
Furthermore, if you want to parse the whole line, you must call parseAll,
otherwise it parses only the beginning of the string in non-greedy manner.
Final remark: for parsing key-value pairs line by line, some String.split and
String.trim would be completely sufficient. Scala Parser Combinators are a little overkill for that.
PS: Hmm... Did you want to allow =-signs in your key-names? Then my version would not work here, because it does not enforce an empty space after the key-name.

Answer 2

This is not a duplicate, it's a different version with RegexParsers that takes care of whitespace explicitly
If you for some reason really care about the white space, then you could stick to the RegexParsers, and do the following (notice the skipWhitespace = false, explicit parser for whitespace ws, the two ws with squiglies around the equality sign, and the repsep with explicitly specified ws):
import scala.util.parsing.combinator._
import scala.io.Source.fromFile
class MyParser extends RegexParsers {
override def skipWhitespace(): Boolean = false
def ws: Parser[String] = "[ \t]+".r
def key: Parser[String] = """[a-zA-Z]+""".r ^^ { _.toString }
def eq: Parser[String] = ws ~> """=""" <~ ws
def string: Parser[String] = """[^ \t\n]+""".r
def value: Parser[List[String]] = repsep(string, ws)
def foo(key: String, value: String): Boolean = {
print(key + " = " + value)
true
}
def parse1: Parser[Boolean] = (key ~ eq ~ string) ^^ { case k ~ e ~ v => foo(k, v) }
def parse2: Parser[Boolean] = (key ~ eq ~ value) ^^ { case k ~ e ~ v => foo(k, v.toString) }
def parseLine(line: String): Boolean = {
parseAll(parse2, line) match {
case Success(matched, _) => true
case Failure(msg, _) => false
case Error(msg, _) => false
}
}
}
val mp = new MyParser()
for (line <- List("hey = hou", "hello = world ppl", "foo = bar baz blup", "foo= bar baz", "foo =bar baz")) {
println(" (Matches: " + mp.parseLine(line) + ")")
}
Now the parser rejects the lines where there is no whitespace around the equal sign:
hey = List(hou) (Matches: true)
hello = List(world, ppl) (Matches: true)
foo = List(bar, baz, blup) (Matches: true)
(Matches: false)
(Matches: false)
The bug with * instead of + in string has been removed, just like in the previous version.