Parsing parentheses on search expression - Scala parser combinator

Question

I am writing a parser for a search expression. Ex.

a = "zyx" and ( b < 5 or c > 9)

I wrote this parser but it's not able to match the parentheses, getting this error:

failure: identifier expected
a = "zyx" and ( b < 5 or c > 9)
              ^

What can I do to be able to match the paretheses

class SearchQueryParser extends StandardTokenParsers {
  def expr: Parser[Expression] = orExp | "(" ~> orExp ~ ")"

  def orExp: Parser[Expression] = {
    andExp *("or" ^^^ {(a: Expression, b: Expression) => BoolExp("OR", (a, b))})
  }

  def andExp: Parser[Expression] = {
    compareExp *("and" ^^^ {(a: Expression, b: Expression) => BoolExp("AND", (a, b))})
  }

  def compareExp: Parser[Expression] = {
    identifier ~ rep(
      ("=" | "!=" | "<" | ">") ~ literal ^^ {
        case op ~ rhs => (op, rhs)
      }
    ) ^^ {
      case (acc, ("=", rhs: Expression)) => Binomial("=", acc, rhs)  
      case (acc, ("!=", rhs: Expression)) => Binomial("!=", acc, rhs)
      case (acc, ("<", rhs: Expression)) => Binomial("<", acc, rhs)
      case (acc, (">", rhs: Expression)) => Binomial(">", acc, rhs)
    }
  }
}

Your compareExp rule looks strange to me. It allows x < 5 > 1, but not 1 < x < 5. I'm not sure whether that's what you intended or not. — sepp2k
– sepp2k, Commented Jul 31, 2019 at 15:05
thanks for your comment. I actually don't need any of the expression as only Binomials should be allowed. I change my grammar accordingly. — locorecto
– locorecto, Commented Jul 31, 2019 at 15:30

sepp2k · Accepted Answer · 2019-07-31 15:05:42Z

1

Your current grammar only allows parentheses in the expr rule, which I assume is your main rule, and the expr rule is never used by any other rule. So parentheses are only allowed around the entire expression.

What you want to do is to put "(" ~ expr ~ ")" in the lowest place where parenthesized expressions would be allowed. As I understand your grammar, that would probably meaning allowing it as an alternative to compareExp in andExp (assuming you don't want to allow parentheses inside compareExps).

answered Jul 31, 2019 at 15:05

sepp2k

372k56 gold badges687 silver badges687 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

locorecto · Accepted Answer · 2019-07-31 15:35:16Z

1

As @sepp2k mentioned you need to put your parentheses on the lowest place where parentheses is allowed. In your case it should be in the compareExp:

  def compareExp: Parser[Expression] = {
    "(" ~> expr <~ ")" | identifier ~ rep(
      ("=" | "!=" | "<" | ">") ~ literal ^^ {
        case op ~ rhs => (op, rhs)
      }
    ) ^^ {
      case (acc, ("=", rhs: Expression)) => Binomial("=", acc, rhs)  
      case (acc, ("!=", rhs: Expression)) => Binomial("!=", acc, rhs)
      case (acc, ("<", rhs: Expression)) => Binomial("<", acc, rhs)
      case (acc, (">", rhs: Expression)) => Binomial(">", acc, rhs)
    }
  }

and exp method should not handle the parentheses

 def expr: Parser[Expression] = orExp

edited Jul 31, 2019 at 15:35

locorecto

1,2033 gold badges14 silver badges42 bronze badges

answered Jul 31, 2019 at 15:28

nolio

838 bronze badges

Collectives™ on Stack Overflow

Parsing parentheses on search expression - Scala parser combinator

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related