You should know that the answers are at: http://en.wikibooks.org/wiki/Write_Yourself_a_Scheme_in_48_Hours/Answers parseNumber'' = many1 digit >>= return . Number . read On Wed, Feb 23, 2011 at 5:31 AM, Chaddaï Fouché wrote:

On Wed, Feb 23, 2011 at 6:30 AM,   wrote:

...

I'm working through the "Write Yourself a Scheme" wikibook and having trouble with one of the exercises.  For the function parseNumber :: Parser LispVal, both

parseNumber = liftM (Number . read) $ many1 digit parseNumber' = do digits <- many1 digit                 return $ (Number . read) digits

work.  But

parseNumber'' = many1 digit >>= liftM read >>= liftM Number

You misunderstand liftM : liftM takes an ordinary function (a -> b) and "lift" it so that it acts inside the monad and become (m a -> m b) thus the parameter of "liftM f" must be a monadic action but through (>>=) you feed it an ordinary value since (>>=) bind a monadic action (m a) to a function that takes an ordinary value and return an action (a -> m b). You used liftM as if it was (return .)

The correct usage would have been :

...

parseNumber'' = liftM (Number . read) (many1 digit)

like your parseNumber

or (using Applicative if you wish to do so) :

...

parseNumber'' = Number . read <$> many1 digit

-- Jedaï

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