On 19 May 2009, at 09:06, Ryan Ingram wrote:
This is a common problem with trying to use do-notation; there are
some cases where you can't make the object an instance of Monad. The
same problem holds for Data.Set; you'd can write
setBind :: Ord b = Set a - (a - Set b) - Set b
setBind m f =
2009/05/18 Miguel Mitrofanov miguelim...@yandex.ru:
On 19 May 2009, at 09:06, Ryan Ingram wrote:
This is a common problem with trying to use do-notation; there are
some cases where you can't make the object an instance of Monad. The
same problem holds for Data.Set; you'd can write
setBind
On Mon, May 18, 2009 at 10:06 PM, Ryan Ingram ryani.s...@gmail.com wrote:
On Mon, May 18, 2009 at 3:05 PM, Taral tar...@gmail.com wrote:
Will this do?
(=) :: (NFData sa, NFData b) = LI sa - (sa - LI b) - LI b
No, the problem is that = on monads has no constraints, it must have the
type
LI
On Mon, 18 May 2009, Nicolas Pouillard wrote:
Excerpts from Jason Dusek's message of Sun May 17 15:45:25 +0200 2009:
From the documentation:
LI could be a strict monad and a strict applicative functor.
However it is not a lazy monad nor a lazy applicative
functor as required
Excerpts from Taral's message of Tue May 19 00:05:39 +0200 2009:
On Mon, May 18, 2009 at 10:30 AM, Nicolas Pouillard
nicolas.pouill...@gmail.com wrote:
The type I would need for bind is this one:
(=) :: NFData sa = LI sa - (sa - LI b) - LI b
Will this do?
(=) :: (NFData sa, NFData
To be fair, you can do this with some extensions; I first saw this in
a paper on Oleg's site [1]. Here's some sample code:
{-# LANGUAGE NoImplicitPrelude, TypeFamilies, MultiParamTypeClasses #-}
module SetMonad where
import qualified Data.Set as S
import qualified Prelude as P (Monad, (=), (),
I've posted it once or twice.
newtype C m r a = C ((a - m r) - m r)
It's a monad, regardless of whether m is one or not. If you have something like return and bind, but not exactly the same, you can make
casting functions
m a - C m r a
and backwards.
Jason Dusek wrote on 19.05.2009 10:23:
Excerpts from Ryan Ingram's message of Tue May 19 10:23:01 +0200 2009:
To be fair, you can do this with some extensions; I first saw this in
a paper on Oleg's site [1]. Here's some sample code:
This seems like the same trick as the rmonad package:
Minor addition, optimize
(I couldn't help myself!)
-- ryan
instance Ord b = ConstrainedBind (S.Set a) (S.Set b) where
type BindElem (S.Set a) = a
m = f = S.unions $ map f $ S.toList m
m n = if S.null m then S.empty else n
___
Nicolas Pouillard wrote:
Excerpts from Ryan Ingram's message of Tue May 19 10:23:01 +0200 2009:
To be fair, you can do this with some extensions; I first saw this in
a paper on Oleg's site [1]. Here's some sample code:
This seems like the same trick as the rmonad package:
On Tue, May 19, 2009 at 12:54 AM, Miguel Mitrofanov
miguelim...@yandex.ru wrote:
I've posted it once or twice.
newtype C m r a = C ((a - m r) - m r)
It's a monad, regardless of whether m is one or not. If you have something
like return and bind, but not exactly the same, you can make casting
Excerpts from Jason Dusek's message of Sun May 17 15:45:25 +0200 2009:
From the documentation:
LI could be a strict monad and a strict applicative functor.
However it is not a lazy monad nor a lazy applicative
functor as required Haskell. Hopefully it is a lazy
(pointed)
On Mon, May 18, 2009 at 10:30 AM, Nicolas Pouillard
nicolas.pouill...@gmail.com wrote:
The type I would need for bind is this one:
(=) :: NFData sa = LI sa - (sa - LI b) - LI b
Will this do?
(=) :: (NFData sa, NFData b) = LI sa - (sa - LI b) - LI b
--
Taral tar...@gmail.com
Please let me
On Mon, May 18, 2009 at 3:05 PM, Taral tar...@gmail.com wrote:
Will this do?
(=) :: (NFData sa, NFData b) = LI sa - (sa - LI b) - LI b
No, the problem is that = on monads has no constraints, it must have the type
LI a - (a - LI b) - LI b
This is a common problem with trying to use
From the documentation:
LI could be a strict monad and a strict applicative functor.
However it is not a lazy monad nor a lazy applicative
functor as required Haskell. Hopefully it is a lazy
(pointed) functor at least.
I'd like to understand this better -- how is LI
On Fri, Mar 20, 2009 at 07:42:28PM +0100, Nicolas Pouillard wrote:
We have good news (nevertheless we hope) for all the lazy guys standing there.
Since their birth, lazy IOs have been a great way to modularly leverage all
the
good things we have with *pure*, *lazy*, *Haskell* functions to the
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