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Today's Topics:
1. Re: How to link two Types (David McBride)
2. Re: How to link two Types (PICCA Frederic-Emmanuel)
3. Re: How to link two Types (PICCA Frederic-Emmanuel)
4. Re: How to link two Types (PICCA Frederic-Emmanuel)
5. Re: Making a Tic-Tac-Toe Game (KC)
6. Re: Understanding the function monad ((->) r) (Michael Litchard)
----------------------------------------------------------------------
Message: 1
Date: Wed, 22 Feb 2017 11:34:43 -0500
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] How to link two Types
Message-ID:
<can+tr42a54aosg6fjmr+bse7db2jayw0jg054er4wwycdkx...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
I should mention I just wrote that code off the cuff. It's probably
not even close to right. I recommend you mess with type families a
little to see if they get you where you want to go in your existing
code.
On Wed, Feb 22, 2017 at 11:29 AM, David McBride <[email protected]> wrote:
> It is hard to tell from your code what you intend, but it works
> however you want it to, so long as it type checks.
>
> class Frame a where
> type Whatever a
> len :: Whatever a -> IO (Maybe Int)
> row :: Whatever a -> MaybeT IO (DifTomoFrame a DIM1)
>
> instance Frame DataFrameH5Path where
> type Whatever DataFrameH5Path = DataFrameH5
> len = undefined -- :: DataFrameH5 -> IO (Maybe Int)
> row = undefined -- :: DataFrameH5 -> Int -> MaybeT (DifTomoFrame
> DataFrameH5Path DIM1)
>
>
>
> On Wed, Feb 22, 2017 at 11:19 AM, PICCA Frederic-Emmanuel
> <[email protected]> wrote:
>> Hello thanks, I will investigate, but I like this solution.
>> I can ad more type to a type family right ?
>>
>>
>> Is it possible with this type family to be able to link in the other way ?
>>
>> a -> t
>>
>> Cheers
>>
>> Fred
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
Message: 2
Date: Wed, 22 Feb 2017 16:40:47 +0000
From: PICCA Frederic-Emmanuel
<[email protected]>
To: "The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell" <[email protected]>
Subject: Re: [Haskell-beginners] How to link two Types
Message-ID:
<a2a20ec3b8560d408356cac2fc148e53bb348...@sun-dag3.synchrotron-soleil.fr>
Content-Type: text/plain; charset="us-ascii"
> It is hard to tell from your code what you intend, but it works
> however you want it to, so long as it type checks.
> class Frame a where
> type Whatever a
> len :: Whatever a -> IO (Maybe Int)
> row :: Whatever a -> MaybeT IO (DifTomoFrame a DIM1)
> instance Frame DataFrameH5Path where
> type Whatever DataFrameH5Path = DataFrameH5
> len = undefined -- :: DataFrameH5 -> IO (Maybe Int)
> row = undefined -- :: DataFrameH5 -> Int -> MaybeT (DifTomoFrame
> DataFrameH5Path DIM1)
In fact what I try realy to do is this.
data DataFrameH5Path
= DataFrameH5Path
(DataItem H5) -- image
(DataItem H5) -- gamma
(DataItem H5) -- delta
(DataItem H5) -- wavelength
deriving (Show)
data DataFrameH5 a
= DataFrameH5
(Nxs a) -- Nexus file
(DataSource H5) -- gamma
(DataSource H5) -- delta
(DataSource H5) -- wavelength
PoniGenerator -- ponie generator
withDataFrameH5 :: (Frame a, MonadSafe m) => File -> Nxs (Key a) ->
PoniGenerator -> (a -> m r) -> m r
withDataFrameH5 h nxs'@(Nxs _ _ (DataFrameH5Path _ g d w)) gen = bracket
(liftIO before) (liftIO . after)
where
-- before :: File -> DataFrameH5Path -> m DataFrameH5
before :: IO a
before = DataFrameH5
<$> return nxs'
<*> openDataSource h g
<*> openDataSource h d
<*> openDataSource h w
<*> return gen
after :: a -> IO ()
after (DataFrameH5 _ g' d' w' _) = do
closeDataSource g'
closeDataSource d'
closeDataSource w'
I open and hdf5 file and I need to read a bunch of data from this file.
the DataFrameH5 is a sort of resource like a File handler.
I need a location in the file in order to acce the data, then I need to close
the file
So I store in the H5 type these resource, that I can release at the end.
Ideally I would like to have only The H5Path type and hide the H5 one but I do
not know how to do this.
I have in fact different H5Path types which necessitate each time there
corresponding H5 type.
So I want a one for one relation between the H5 <-> H5Path type.
Cheers
Frederic
------------------------------
Message: 3
Date: Wed, 22 Feb 2017 16:41:19 +0000
From: PICCA Frederic-Emmanuel
<[email protected]>
To: "The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell" <[email protected]>
Subject: Re: [Haskell-beginners] How to link two Types
Message-ID:
<a2a20ec3b8560d408356cac2fc148e53bb348...@sun-dag3.synchrotron-soleil.fr>
Content-Type: text/plain; charset="us-ascii"
> I should mention I just wrote that code off the cuff. It's probably
> not even close to right. I recommend you mess with type families a
> little to see if they get you where you want to go in your existing
> code.
I am playing with it thanks :))
------------------------------
Message: 4
Date: Wed, 22 Feb 2017 17:19:10 +0000
From: PICCA Frederic-Emmanuel
<[email protected]>
To: "The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell" <[email protected]>
Subject: Re: [Haskell-beginners] How to link two Types
Message-ID:
<a2a20ec3b8560d408356cac2fc148e53bb349...@sun-dag3.synchrotron-soleil.fr>
Content-Type: text/plain; charset="us-ascii"
Hello, I am affected by this error
src/Hkl/Xrd/OneD.hs:238:49-52:
Could not deduce (Key a ~ Key b0)
from the context (Frame a)
bound by the type signature for
getPoniExtRef :: Frame a => XRDRef (Key a) -> IO PoniExt
at src/Hkl/Xrd/OneD.hs:235:18-56
NB: `Key' is a type function, and may not be injective
The type variable `b0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Expected type: Nxs (Key b0)
Actual type: Nxs (Key a)
In the second argument of `withDataFrameH5', namely nxs'
In the first argument of `(>->)', namely
`withDataFrameH5 h5file nxs' (gen output f) yield'
In the first argument of `toListM', namely
`(withDataFrameH5 h5file nxs' (gen output f) yield
>-> hoist lift (frames' [idx]))'
I looked at this
http://stackoverflow.com/questions/20870432/type-family-vs-data-family-in-brief-haskell#20908500
So in your opinion it would be better to use a data family instead of a type
familly ?
Cheers
------------------------------
Message: 5
Date: Wed, 22 Feb 2017 10:14:50 -0800
From: KC <[email protected]>
To: Haskell Beginners <[email protected]>
Subject: Re: [Haskell-beginners] Making a Tic-Tac-Toe Game
Message-ID:
<CAMLKXymxsTjxypm0Hnr7b2sO=ztm86jtgx-qiknsacbfsgr...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
If you want graphics there is
Making your first Haskell game
Which uses hGamer3D and aio
--
--
Sent from an expensive device which will be obsolete in a few months! :D
Casey
On Feb 21, 2017 8:36 AM, "Sudhanshu Jaiswal" <[email protected]>
wrote:
> Also called as Noughts and crosses or Xs and Os.
>
> Hello everyone,
>
> How do I start making a *Two Player* Tic Tac Toe game in Haskell?
>
> The program basically has to show the 3*3 grid as coordinates and let each
> player choose his coordinates in his turn by entering the coordinates of
> the required cell. I also want to be able to check if a player has won and
> display it once the winning move has been made or the same for a Draw.
>
> What have I done? - I have read Learn You a Haskell till Modules and know
> basic I/O.
>
> I don't want the code instead, I am interested in learning stuff and
> trying problems which would lead me to get the intuition and ability to
> make the game by myself.
>
> I would be thankful if you folks could direct me to related problems which
> I could do or some advice as to how I should go about implementing such a
> program.
> --
> Sudhanshu
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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Message: 6
Date: Wed, 22 Feb 2017 17:50:05 -0800
From: Michael Litchard <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Understanding the function monad
((->) r)
Message-ID:
<caezekyrgn5k7q-leoeh3-rv3najgnv71fzg+dbyp9fvodda...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Thanks for this response. the Monad instance for ((->) r) has been bugging
me as well.
On Tue, Feb 21, 2017 at 6:32 AM, Rahul Muttineni <[email protected]>
wrote:
> Hi Olumide,
>
> Let the types help you out.
>
> The Monad typeclass (omitting the superclass constraints):
>
> class Monad m where
> return :: a -> m a
> (>>=) :: m a -> (a -> m b) -> m b
>
> Write out the specialised type signatures for (->) r:
>
> {-# LANGUAGE InstanceSigs #-}
> -- This extension allows you to specify the type signatures in instance
> declarations
>
> instance Monad ((->) r) where
> return :: a -> (r -> a)
> (>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b)
>
> Now we look at how to make some definition of return that type checks.
> We're given an a and we want to return a function that takes an r and
> returns an a. Well the only way you can really do this is ignoring the r
> and returning the value you were given in all cases! Because 'a' can be
> *anything*, you really don't have much else you can do! Hence:
>
> return :: a -> (r -> a)
> return a = \_ -> a
>
> Now let's take a look at (>>=). Since this is a bit complicated, let's
> work backwards from the result type. We want a function that gives us a b
> given an r and given two functions with types (r -> a) and (a -> (r -> b)).
> To get a b, we need to use the second function. To use the second function,
> we must have an a, which we can get from the first function!
>
> (>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b)
> (>>=) f g = \r -> (g (f r)) r
>
> Hope that helps!
> Rahul
>
>
> On Tue, Feb 21, 2017 at 5:04 PM, Olumide <[email protected]> wrote:
>
>> On 21/02/2017 10:25, Benjamin Edwards wrote:
>>
>>> What is it that you are having difficulty with? Is it "why" this is a
>>> good definition? Is it that you don't understand how it works?
>>>
>>
>> I simply can't grok f (h w) w.
>>
>> - Olumide
>>
>> On Tue, 21 Feb 2017 at 10:15 Olumide <[email protected]
>>> <mailto:[email protected]>> wrote:
>>>
>>> Hello List,
>>>
>>> I am having enormous difficulty understanding the definition of the
>>> bind
>>> operator of ((->) r) as show below and would appreciate help i this
>>> regard.
>>>
>>> instance Monad ((->) r) where
>>> return x = \_ -> x
>>> h >>= f = \w -> f (h w) w
>>>
>>> Thanks,
>>>
>>> - Olumide
>>>
>>> _______________________________________________
>>> Beginners mailing list
>>> [email protected] <mailto:[email protected]>
>>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>>
>>>
>>>
>>> _______________________________________________
>>> Beginners mailing list
>>> [email protected]
>>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>>
>>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>
>
>
> --
> Rahul Muttineni
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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