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 complains "couldn't match expected type Char against inferred type [Char]". Is liftM engaging the list monad? How do I fix this?
On Wed, Feb 23, 2011 at 6:30 AM, <blackcat@pro-ns.net> 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
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 <chaddai.fouche@gmail.com> wrote:
On Wed, Feb 23, 2011 at 6:30 AM, <blackcat@pro-ns.net> 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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