What's the best way to write a parser by hand? [closed]

Question

We've used ANTLR to create a parser for a SQL-like grammar, and while the results are satisfactory in most cases, there are a few edge cases that we need to fix; and since we didn't write the parser ourselves we don't really understand it well enough to be able to make sensible changes.

So, we'd like to write our own parser. What's the best way to go about writing a parser by hand? What sort of parser should we use - recursive descent has been recommended; is that right? We'll be writing it in C#, so any tutorials for writing parsers in that language would be gratefully received.

UPDATE: I'd also be interested in answers that involve F# - I've been looking for a reason to use that in a project.

Answer 1

I would highly recommend the F# language as your language of choice for parsing on the .NET Platform. It's roots in the ML family of languages means it has excellent support for language-oriented programming.

Discriminated unions and pattern-matching allow for a very succinct and powerful specification of your AST. Higher-order functions allow for definition of parse operations and their composition. First-class support for monadic types allows for state management to be handled implicitly greatly simplifying the composition of parsers. Powerful type-inference greatly aides the definition of these (complex) types. And all of this can be specified and executed interactively allowing you to rapidly prototype.

Stephan Tolksdorf has put this into practice with his parser combinator library FParsec

From his examples we see how naturally an AST is specified:

type expr =
    | Val of string
    | Int of int
    | Float of float
    | Decr of expr

type stmt =
    | Assign of string * expr
    | While of expr * stmt
    | Seq of stmt list
    | IfThen of expr * stmt
    | IfThenElse of expr * stmt * stmt
    | Print of expr

type prog = Prog of stmt list

the implementation of the parser (partially elided) is just as succinct:

let stmt, stmtRef = createParserForwardedToRef()

let stmtList = sepBy1 stmt (ch ';')

let assign =
    pipe2 id (str ":=" >>. expr) (fun id e -> Assign(id, e))

let print = str "print" >>. expr |>> Print

let pwhile =
    pipe2 (str "while" >>. expr) (str "do" >>. stmt) (fun e s -> While(e, s))

let seq =
    str "begin" >>. stmtList .>> str "end" |>> Seq

let ifthen =
    pipe3 (str "if" >>. expr) (str "then" >>. stmt) (opt (str "else" >>. stmt))
          (fun e s1 optS2 ->
               match optS2 with
               | None    -> IfThen(e, s1)
               | Some s2 -> IfThenElse(e, s1, s2))

do stmtRef:= choice [ifthen; pwhile; seq; print; assign]


let prog =
    ws >>. stmtList .>> eof |>> Prog

On the second line, as an example, stmt and ch are parsers and sepBy1 is a monadic parser combinator that takes two simple parsers and returns a combination parser. In this case sepBy1 p sep returns a parser that parses one or more occurrences of p separated by sep. You can thus see how quickly a powerful parser can be combined from simple parsers. F#'s support for overridden operators also allow for concise infix notation e.g. the sequencing combinator and the choice combinator can be specified as >>. and <|>.

Best of luck,

Danny

Answer 2

The only kind of parser that can be handwritten by a sane human being is a recursive-descent. That said, writing bottom-up parser by hand is still possible but is very undesirable.

If you're up for RD parser you have to verify that SQL grammar is not left-recursive (and eliminate the recursion if necessary), and then basically write a function for each grammar rule. See this for further reference.

Answer 3

Adding my voice to the chorus in favor of recursive-descent (LL1). They are simple, fast, and IMO, not at all hard to maintain.

However, take a good look at your language to make sure it is LL1. If you have any syntax like C has, like ((((type))foo)[ ]) where you might have to descend multiple layers of parentheses before you even find out if you are looking at a type, variable, or expression, then LL1 will be very difficult, and bottom-up wins.