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
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@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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