Re: [Haskell-cafe] i am missing something really trivial with parsec

I don't know, but: number -- definition = do { num <- natural ; return $ num } -- desugar = natural >>= \num -> return $ num -- apply ($) = natural >>= \num -> return num -- eta elimination (f == \x -> f x) = natural >>= return -- monad law = natural (modulo monomorphism restriction, since number doesn't take any arguments and doesn't have a type signature) -- ryan On Tue, Sep 29, 2009 at 12:54 AM, Anatoly Yakovenko wrote:

number = do { num <- natural ; return $ num } main = do txt <- hGetContents stdin print $ parse number "stdin" txt

why doesn't that work? _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe