Hi,
Haskell expects the function with type (a -> m b) in the right side of
(>>=),
but you put there function with type (a -> a):
try:
:t (Just 3 >>=)
(Just 3 >>=) :: (Num a) => (a -> Maybe b) -> Maybe b
and:
:t (1+)
(1+) :: (Num a) => a -> a
You should put (1+) into Maybe monad, just do return.(1+), so
Just 3 >>= return . (1+)
will return `Just 4`
--
Best regards,
Vasyl Pasternak
2009/5/9 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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