[Haskell-beginners] About monad

[Trung Quang Nguyen]

*fmap*<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:fmap>
::
Functor f => (a -> b) -> f a -> f
b<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:fmap>
fmap f (Just a)      = Just (f a)

We wrap Just around (f a) because f return a value with type b instead
(Just b).

But in
(*>>=*)<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:-62--62--61->
::
Monad m => m a -> (a -> m b) -> m
b<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:-62--62--61->
Just x >>= f  = f x

We don't need to wrap Just around (f a) because f return (Just b).

--Trung



2012/12/20 Miguel Negrao <miguel.negrao-lists at friendlyvirus.org>

>
> A 20/12/2012, às 14:07, Trung Quang Nguyen escreveu:
>
> > Hi all,
> >
> > I saw this
> >
> >       • instance Monad Maybe where
> >       •     return x = Just x
> >       •     Nothing >>= f = Nothing
> >       •     Just x >>= f  = f x
> >       •     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?
>
> That would be the implementation of fmap for Maybe:
>
> instance  Functor Maybe  where
>     fmap _ Nothing       = Nothing
>     fmap f (Just a)      = Just (f a)
>
> so, different behavior.
>
> best,
> Miguel
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-- 
*Trung Nguyen*
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