Backtracking in scala parser combinators?

Question

It seems like scala's parser combinators don't backtrack. I have a grammar (see bottom) which can't parse the following "stmt" correctly:

copy in to out .

That should be easy to parse with backtracking:

stmt: (to out(copy in))

or am I missing something?

Parser:

type ExprP = Parser[Expr]
type ValueP = Parser[ValExpr]
type CallP = Parser[Call]
type ArgsP = Parser[Seq[Expr]]

val ident     = "[a-zA-Z\\+\\-\\*/%><\\\\\\=]+".r
val sqstart   = "\\["                          .r
val sqend     = "\\]"                          .r
val del       = ","                            .r
val end       = "\\."                          .r

def stmt: ExprP      = expr <~ end
def expr: ExprP      = ucall | call | value
def value: ValueP    = ident ^^ {str => IdentExpr(str)}
def call: CallP      = (args ~ ident ~ expr) ^^ {case args ~ method ~ upon => Call(args, method, upon)}
def ucall: CallP     = (ident ~ expr) ^^ {case method ~ upon => Call(Seq(), method, upon)}
def args: ArgsP      = advargs | smplargs
def smplargs: ArgsP  = expr ^^ {e => Seq(e)}
def advargs: ArgsP   = (sqstart ~> repsep(expr, del) <~ sqend) ^^ {seq => seq}

Answer 1

You want to use PackratParsers in 2.8. I think the packrat parser is the only backtracking parser.

Edit: as of mid-year 2015, you should use fastparse instead. It's not only much faster, but also easier to use (especially when building data structures from parsing).

Answer 2

Your problem is not backtracking. The standard | operator in scala.util.parsing.combinator will do backtracking. Your problem is left-recursion ( expr → call → args → smplargs → expr ). Packrat parsing may indeed help with that.