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Today's Topics:
1. About monad (Trung Quang Nguyen)
2. Re: About monad (Edward Z. Yang)
3. Re: About monad (Tom Davie)
4. Re: About monad (Trung Quang Nguyen)
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Message: 1
Date: Thu, 20 Dec 2012 15:07:02 +0100
From: Trung Quang Nguyen <[email protected]>
Subject: [Haskell-beginners] About monad
To: beginners <[email protected]>
Message-ID:
<CALs5uBy97og=dse2wFAObVCw=t8H0Xi6r985ar95Fne=fp+...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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?
--Trung
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Message: 2
Date: Thu, 20 Dec 2012 22:12:02 +0800
From: "Edward Z. Yang" <[email protected]>
Subject: Re: [Haskell-beginners] About monad
To: Trung Quang Nguyen <[email protected]>
Cc: beginners <[email protected]>
Message-ID: <1356012708-sup-2809@javelin>
Content-Type: text/plain; charset=UTF-8
Excerpts from Trung Quang Nguyen's message of Thu Dec 20 22:07:02 +0800 2012:
> 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)*?
Try it; it won't typecheck.
Cheers,
Edward
------------------------------
Message: 3
Date: Thu, 20 Dec 2012 14:13:28 +0000
From: Tom Davie <[email protected]>
Subject: Re: [Haskell-beginners] About monad
To: Trung Quang Nguyen <[email protected]>
Cc: beginners <[email protected]>
Message-ID: <[email protected]>
Content-Type: text/plain; charset="us-ascii"
On 20 Dec 2012, at 14:07, Trung Quang Nguyen <[email protected]> wrote:
> 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?
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
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Message: 4
Date: Thu, 20 Dec 2012 15:25:45 +0100
From: Trung Quang Nguyen <[email protected]>
Subject: Re: [Haskell-beginners] About monad
To: Tom Davie <[email protected]>
Cc: beginners <[email protected]>
Message-ID:
<CALs5uBxEJJxM-CEqw5RxQKjByqEgRxeDp=cqqvxhl0w8aed...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <[email protected]>
>
> On 20 Dec 2012, at 14:07, Trung Quang Nguyen <[email protected]> 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/
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