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 <[email protected]> > 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 > > ================ > > [mich...@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> > > > > _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe > >
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