[Haskell-cafe] Just 3 >>= (1+)?

[Bas van Gijzel]

Hey Michael,

If you would look at the type of  >>=, it would give
(>>=) :: (Monad m) => m a -> (a -> m b) -> m b

and specifically in your case:
(>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b

You are applying Just 3 as first argument, which is correct, but then supply
a partially applied function (1+) which is of type Num  a => a -> a, while
it should be
a -> Maybe b.

What are you expecting as result? You cannot pull something out of a monad
using a bind operator. Maybe you meant something like this?
(Just 3) >>= \x -> return (x + 1)
Notice how Just 3 is just the Maybe a argument, and \x -> return (x + 1) is
the (a -> Maybe b) argument, finally delivering a Just 4 (of type Maybe b).

(This is the same as do x <- Just 3
                                   return (x + 1)
)


Oh and btw, fail should take an argument (the error string).

Good luck,

Bas van Gijzel

On Sat, May 9, 2009 at 9:31 PM, michael rice <nowgate at yahoo.com> wrote:

> Why doesn't this work?
>
> Michael
>
> ================
>
> data Maybe a = Nothing | Just a
>
> instance Monad Maybe where
>     return         = Just
>     fail           = Nothing
>     Nothing  >>= f = Nothing
>     (Just x) >>= f = f x
>
> instance MonadPlus Maybe where
>     mzero             = Nothing
>     Nothing `mplus` x = x
>     x `mplus` _       = x
>
> ================
>
> [michael at localhost ~]$ ghci
> GHCi, version 6.10.1: http://www.haskell.org/ghc/  :? for help
> Loading package ghc-prim ... linking ... done.
> Loading package integer ... linking ... done.
> Loading package base ... linking ... done.
> Prelude> Just 3 >>= (1+)
>
> <interactive>:1:0:
>     No instance for (Num (Maybe b))
>       arising from a use of `it' at <interactive>:1:0-14
>     Possible fix: add an instance declaration for (Num (Maybe b))
>     In the first argument of `print', namely `it'
>     In a stmt of a 'do' expression: print it
> Prelude>
>
>
>
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