Backtracking in scala parser combinators?
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26-09-2019 - |
문제
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}
해결책
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).
다른 팁
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.
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