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Today's Topics:
1. Re: How to create a monad in GHC 7.10 or newer (Ahmad Ismail)
2. Re: How to create a monad in GHC 7.10 or newer (Francesco Ariis)
3. Re: How to create a monad in GHC 7.10 or newer (Ahmad Ismail)
----------------------------------------------------------------------
Message: 1
Date: Sun, 13 Nov 2022 20:56:32 +0600
From: Ahmad Ismail <ismail...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] How to create a monad in GHC 7.10 or
newer
Message-ID:
<CAHAhJwJBDRMy1wuPqSHVHUVJHcVQ-e4TcWk7k5d=CH+=scv...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
How can I fix it so that `ItDoesnt <*> WhatThisIsCalled` works?
I have came up with a solution without WhatThisIsCalled
data WhoCares a = ItDoesnt | Matter a deriving (Eq, Show)
instance Functor WhoCares where
fmap _ ItDoesnt = ItDoesnt
fmap f (Matter a) = Matter (f a)
instance Applicative WhoCares where
pure = Matter
Matter f <*> Matter a = Matter (f a)
ItDoesnt <*> _ = ItDoesnt
_ <*> ItDoesnt = ItDoesnt
instance Monad WhoCares where
return x = Matter x
(Matter x) >>= k = k x
ItDoesnt >>= _ = ItDoesnt
half x = if even x
then Matter (x `div` 2)
else ItDoesnt
incVal :: (Ord a, Num a) => a -> WhoCares a
incVal x
| x + 1 <= 10 = return (x + 1)
| otherwise = ItDoesnt
decVal :: (Ord a, Num a) => a -> WhoCares a
decVal x
| x - 1 >= 0 = return (x - 1)
| otherwise = ItDoesnt
main = do
-- fmap id == id
let funcx = fmap id "Hi Julie"
let funcy = id "Hi Julie"
print(funcx)
print(funcy)
print(funcx == funcy)
-- fmap (f . g) == fmap f . fmap g
let funcx' = fmap ((+1) . (*2)) [1..5]
let funcy' = fmap (+1) . fmap (*2) $ [1..5]
print(funcx')
print(funcy')
print(funcx' == funcy')
-- pure id <*> v = v
print(pure id <*> (Matter 10))
-- pure (.) <*> u <*> v <*> w = u <*> (v <*> w)
let appx = pure (.) <*> (Matter (+1)) <*> (Matter (*2)) <*> (Matter 10)
let appy = (Matter (+1)) <*> ((Matter (*2)) <*> (Matter 10))
print(appx)
print(appy)
print(appx == appy)
-- pure f <*> pure x = pure (f x)
let appx' = pure (+1) <*> pure 1 :: WhoCares Int
let appy' = pure ((+1) 1) :: WhoCares Int
print(appx')
print(appy')
print(appx' == appy')
-- u <*> pure y = pure ($ y) <*> u
let appx'' = Matter (+2) <*> pure 2
let appy'' = pure ($ 2) <*> Matter (+ 2)
print(appx'')
print(appy'')
print(appx'' == appy'')
-- m >>= return = m
let monx = Matter 20 >>= return
let mony = Matter 20
print(monx)
print(mony)
print(monx == mony)
-- return x >>= f = f x
let monx' = return 20 >>= half
let mony' = half 20
print(monx')
print(mony')
print(monx' == mony')
-- (m >>= f) >>= g = m >>= (\x -> f x >>= g)
let monx'' = return 20 >>= half >>= half
let mony'' = half 20 >>= half
print(monx'')
print(mony'')
print(monx'' == mony'')
print (Matter 7 >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal >>= decVal
>>= decVal)
print (Matter 2 >>= decVal >>= decVal >>= decVal)
print (Matter 20 >>= half >>= half)
*Thanks and Best Regards,Ahmad Ismail*
On Sun, Nov 13, 2022 at 5:08 PM Francesco Ariis <fa...@ariis.it> wrote:
> Hello Ahmad,
>
> Il 13 novembre 2022 alle 16:33 Ahmad Ismail ha scritto:
> > Due to lack of examples, I am not understanding how to implement >>= and
> > >>.
>
> All you need to implement is (>>=)!
>
> > The code I came up with so far is:
> >
> > instance Monad (WhoCares a) where
> > (>>=) :: Matter a -> (a -> Matter b) -> Matter b
> > (>>) :: Matter a -> Matter b -> Matter b
> > return :: a -> Matter a
> > return = pure
>
> The signature for (>>=) is wrong, `Matter` is a *data* constructor, you
> need a *type* one instead, so:
>
> (>>=) :: WhoCares a -> (a -> WhoCares b) -> WhoCares b
>
> But let us go back to typeclasses. Your `Applicative` instance
>
> > instance Applicative WhoCares where
> > pure = Matter
> > Matter f <*> Matter a = Matter (f a)
>
> is broken:
>
> λ> ItDoesnt <*> WhatThisIsCalled
> *** Exception: /tmp/prova.hs:11:5-40: Non-exhaustive patterns in
> function <*>
>
> So we need first to fix that. What behaviour would you expect, what are
> you trying to model with `WhoCares`?
> —F
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 2
Date: Sun, 13 Nov 2022 16:21:49 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] How to create a monad in GHC 7.10 or
newer
Message-ID: <y3eljsmfplgda...@mkiii.casa>
Content-Type: text/plain; charset=us-ascii
Il 13 novembre 2022 alle 20:56 Ahmad Ismail ha scritto:
> How can I fix it so that `ItDoesnt <*> WhatThisIsCalled` works?
>
> I have came up with a solution without WhatThisIsCalled
>
> data WhoCares a = ItDoesnt | Matter a deriving (Eq, Show)
>
> instance Functor WhoCares where
> fmap _ ItDoesnt = ItDoesnt
> fmap f (Matter a) = Matter (f a)
>
> instance Applicative WhoCares where
> pure = Matter
> Matter f <*> Matter a = Matter (f a)
> ItDoesnt <*> _ = ItDoesnt
> _ <*> ItDoesnt = ItDoesnt
>
> instance Monad WhoCares where
> return x = Matter x
> (Matter x) >>= k = k x
> ItDoesnt >>= _ = ItDoesnt
This makes much more sense. Now <*> is total and >>= is meaningful,
well done!
------------------------------
Message: 3
Date: Sun, 13 Nov 2022 21:47:33 +0600
From: Ahmad Ismail <ismail...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] How to create a monad in GHC 7.10 or
newer
Message-ID:
<cahahjwjiq-2-4hpqdzo_rxwiypnodhss8q10tyqm1ow+gia...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Thank you very much.
*Thanks and Best Regards,Ahmad Ismail*
On Sun, Nov 13, 2022 at 9:22 PM Francesco Ariis <fa...@ariis.it> wrote:
> Il 13 novembre 2022 alle 20:56 Ahmad Ismail ha scritto:
> > How can I fix it so that `ItDoesnt <*> WhatThisIsCalled` works?
> >
> > I have came up with a solution without WhatThisIsCalled
> >
> > data WhoCares a = ItDoesnt | Matter a deriving (Eq, Show)
> >
> > instance Functor WhoCares where
> > fmap _ ItDoesnt = ItDoesnt
> > fmap f (Matter a) = Matter (f a)
> >
> > instance Applicative WhoCares where
> > pure = Matter
> > Matter f <*> Matter a = Matter (f a)
> > ItDoesnt <*> _ = ItDoesnt
> > _ <*> ItDoesnt = ItDoesnt
> >
> > instance Monad WhoCares where
> > return x = Matter x
> > (Matter x) >>= k = k x
> > ItDoesnt >>= _ = ItDoesnt
>
> This makes much more sense. Now <*> is total and >>= is meaningful,
> well done!
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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