[Haskell-cafe] Parsing words with parsec

Paolo Veronelli paolo.veronelli at gmail.com
Sat Mar 31 11:10:04 EDT 2007

On Friday 30 March 2007 06:59, Stefan O'Rear wrote:

> Anyway, I think parsec is *far* too big a hammer for the nail you're trying
> to hit.

In the end , the big hammer solution has become

parseLine = fmap (map fst. filter snd) $ many parser 
  where parser = do w <- option ("",False) parseAWord  
                    anyChar -- skip the separator
                    return w
        parseAWord = try positive <|> (many1 nonSeparator >> return 
        positive = do c <- wordChar
                      (cs,tn) <- option ("",True) parseAWord
                      return (c:cs,tn)
wordChar = letter <|> oneOf "_@" <?> "a word-character"
nonSeparator = wordChar <|> digit <?> "a non-separator"

while your, corrected not parsec solution is

wordsOfLine isNonSeparator isWordChar = (filter (all isWordChar)).
     groupBy (\x y -> (isNonSeparator x) == (isNonSeparator y)) 

Still ,I wonder if the parsec solution can be simplified.


(PS. I put an option on the ML software which sends me an ack on posting , so 
at least I know I sent the mail :) )

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