Re: [Haskell-beginners] Re: parsing upto n items with parsec

Right. I took "option" out for some reason. The following definitions have the desired behavior. I also added a base case to fromTo because otherwise the call to count would consume m tokens even if n-m < 0, which is probably not what we'd expect. upTo :: Int -> GenParser tok st a -> GenParser tok st [a] upTo n p | n <= 0 = return [] | otherwise = option [] $ liftM2 (:) p (upTo (n-1) p) fromTo :: Int -> Int -> GenParser tok st a -> GenParser tok st [a] fromTo m n p | n-m < 0 = return [] | otherwise = liftM2 (++) (count m p) (upTo (n-m) p) Thanks! On Mon, Oct 19, 2009 at 12:55 PM, Chaddaï Fouché wrote:

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The semantics of (upTo n p) should be to parse at most n tokens, but if less than n tokens are available that should still be a successful parse. And

On Mon, Oct 19, 2009 at 6:22 PM, Ashish Agarwal wrote: the

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next token should be the first one upTo failed on. I attempted to use the "try" parser in various locations but that doesn't seem to help, or maybe I'm using it incorrectly.

First, you should probably use pattern matching rather than if for your base case/recursive case distinction, it's clearer (at least most Haskellers seems to think it is) :

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upTo 0 p = return []

second, option was conceived for these case where you want to try a parser and returns something if it fail :

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upTo n p = option [] $ liftM2 (:) p (upTo (n-1) p)

Even then, be careful of putting a try before any multi-token parser that could fail harmlessly during the upTo.

-- Jedaï