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

10 May 2009


      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 <nowgate@yahoo.com>
...
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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