[Haskell-cafe] Just 3 >>= (1+)?
Bas van Gijzel
nenekotan at gmail.com
Sat May 9 16:46:57 EDT 2009
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).
Bas van Gijzel
On Sat, May 9, 2009 at 9:31 PM, michael rice <nowgate at yahoo.com> wrote:
> Why doesn't this work?
> 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+)
> 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
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
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