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parsinghaskellparsec

Using Parsec to write a Read instance

Using Parsec, I'm able to write a function of type String -> Maybe MyType with relative ease. I would now like to create a Read instance for my type based on that; however, I don't understand how readsPrec works or what it is supposed to do.

My best guess right now is that readsPrec is used to build a recursive parser from scratch to traverse a string, building up the desired datatype in Haskell. However, I already have a very robust parser who does that very thing for me. So how do I tell readsPrec to use my parser? What is the "operator precedence" parameter it takes, and what is it good for in my context?

If it helps, I've created a minimal example on Github. It contains a type, a parser, and a blank Read instance, and reflects quite well where I'm stuck.

(Background: The real parser is for Scheme.)

Solution

  • However, I already have a very robust parser who does that very thing for me.

    It's actually not that robust, your parser has problems with superfluous parentheses, it won't parse

    ((1) (2))

    for example, and it will throw an exception on some malformed inputs, because

    singleP = Single . read <$> many digit

    may use read "" :: Int.

    That out of the way, the precedence argument is used to determine whether parentheses are necessary in some place, e.g. if you have

    infixr 6 :+:

data a :+: b = a :+: b

data C = C Int

data D = D C

    you don't need parentheses around a C 12 as an argument of (:+:), since the precedence of application is higher than that of (:+:), but you'd need parentheses around C 12 as an argument of D.

    So you'd usually have something like

    readsPrec p = needsParens (p >= precedenceLevel) someParser

    where someParser parses a value from the input without enclosing parentheses, and needsParens True thing parses a thing between parentheses, while needsParens False thing parses a thing optionally enclosed in parentheses [you should always accept more parentheses than necessary, ((((((1)))))) should parse fine as an Int].

    Since the readsPrec p parsers are used to parse parts of the input as parts of the value when reading lists, tuples etc., they must return not only the parsed value, but also the remaining part of the input.

    With that, a simple way to transform a parsec parser to a readsPrec parser would be

    withRemaining :: Parser a -> Parser (a, String)
withRemaining p = (,) <$> p <*> getInput

parsecToReadsPrec :: Parser a -> Int -> ReadS a
parsecToReadsPrec parsecParser prec input
    = case parse (withremaining $ needsParens (prec >= threshold) parsecParser) "" input of
        Left _ -> []
        Right result -> [result]

    If you're using GHC, it may however be preferable to use a ReadPrec / ReadP parser (built using Text.ParserCombinators.ReadP[rec]) instead of a parsec parser and define readPrec instead of readsPrec.