[Haskell-cafe] Confusing `bind` and (>>=)

[Ivan Lazar Miljenovic]

On 10 October 2017 at 22:34, mirone <saul-mirone at qq.com> wrote:
> Hi there,
>   I'm learning haskell for weeks and I'm trying to write a parser using
> haskell.
>   First I create a datatype:
>     newtype Parser a = P (String -> [(a, String)])
>   I create a function called `bind` :
>     bind :: Parser a -> ( a -> Parser b) -> Parser b
>     bind f g = P $ \s -> concatMap (\(a, s') -> parse (g a) s') $ parse f s
>   and then:
>     instance Monad Parser where
>       (>>=)  = bind
>   that's looks cool, than I write a function
>     liftToken :: (Char -> [a]) -> Parser a
>     liftToken f = P g where
>       g [] = []
>       g (c:cs) = f c >>= (\x -> return (x, cs))

Here, f c :: [a], so you're using the [] instance of Monad, hence
(>>=) :: [a] -> (a -> [b]) -> [b]

>   It works well, then I change this function to
>     liftToken :: (Char -> [a]) -> Parser a
>     liftToken f = P g where
>       g [] = []
>       g (c:cs) = f c `bind` (\x -> return (x, cs))
>   GHC throw errors:
>     regular.hs|153 col 14 error| • Couldn't match expected type ‘Parser t0’
> with actual type ‘[a]’ • In the first argument of ‘bind’, namely ‘f c’ In
> the expression: f c `bind` (\ x -> return (x, cs)) In an equation for ‘g’: g
> (c : cs) = f c `bind` (\ x -> return (x, cs)) • Relevant bindings include f
> :: Char -> [a] (bound at regular.hs:151:11) liftToken :: (Char -> [a]) ->
> Parser a (bound at regular.hs:151:1)
>
>   That's really confusing, why I can use (>>=) but I can't use `bind`? Is
> there any difference between them?
>
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-- 
Ivan Lazar Miljenovic
Ivan.Miljenovic at gmail.com
http://IvanMiljenovic.wordpress.com