Re: [Haskell-beginners] About monad

Oh yes, I understand now. Just x >>= f = f x the output of f is actually (Monad value) like in this example (Just 3) >>= (\x -> Just $ x^2) At the first sight, I thought about (Monad (f x)), but it's wrong because it will be (Monad (Monad value)) when f return. Thanks a lot! --Trung 2012/12/20 Tom Davie

On 20 Dec 2012, at 14:07, Trung Quang Nguyen wrote:

Hi all,

I saw this

1. instance Monad Maybe where 2. return x = Just x 3. Nothing >>= f = Nothing 4. Just x >>= f = f x 5. fail _ = Nothing

I am wondering about the implementation of function (>>=). Why don't it be *Just x >>= f = Just (f x)*?

Any body knows about this?

The reason is in the type of bind:

(>>=) :: Monad m => m a -> (a -> m b) -> m b

The function f takes a non-in-a-monad value, and gives you an in-a-monad value.

Bob

-- *Trung Nguyen* Mobile: +45 50 11 10 63 LinkedIn: http://www.linkedin.com/pub/trung-nguyen/36/a44/187 View my blog at http://www.onextrabit.com/