Send Beginners mailing list submissions to
        beginners@haskell.org

To subscribe or unsubscribe via the World Wide Web, visit
        http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
or, via email, send a message with subject or body 'help' to
        beginners-requ...@haskell.org

You can reach the person managing the list at
        beginners-ow...@haskell.org

When replying, please edit your Subject line so it is more specific
than "Re: Contents of Beginners digest..."


Today's Topics:

   1.  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)


----------------------------------------------------------------------

Message: 1
Date: Sun, 13 Nov 2022 16:33:54 +0600
From: Ahmad Ismail <ismail...@gmail.com>
To: beginners@haskell.org
Subject: [Haskell-beginners] How to create a monad in GHC 7.10 or
        newer
Message-ID:
        <CAHAhJwKAnzZ_tGUi81brvUZ8DM+QyJC=irpt_u0oumus8cu...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

In the book "Haskell programming from first principles" it is said that:

If you are using GHC 7.10 or newer, you’ll see an Applicative constraint in
> the definition of Monad, as it should be:



    class Applicative m => Monad m where
>         (>>=) :: m a -> (a -> m b) -> m b
>         (>>) :: m a -> m b -> m b
>         return :: a -> m a


I have created the following applicative functor.

data WhoCares a = ItDoesnt | Matter a | WhatThisIsCalled deriving (Eq, Show)

instance Functor WhoCares where
    fmap _ ItDoesnt = ItDoesnt
    fmap _ WhatThisIsCalled = WhatThisIsCalled
    fmap f (Matter a) = Matter (f a)

instance Applicative WhoCares where
    pure = Matter
    Matter f <*> Matter a = Matter (f a)

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'')

Due to lack of examples, I am not understanding how to implement >>= and
>>. 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

So, that I can do stuff like:

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
    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)

With Output:

10
ItDoesnt
ItDoesnt
ItDoesnt
5

Please help.

*Thanks and Best Regards,Ahmad Ismail*
-------------- next part --------------
An HTML attachment was scrubbed...
URL: 
<http://mail.haskell.org/pipermail/beginners/attachments/20221113/b0dec940/attachment-0001.html>

------------------------------

Message: 2
Date: Sun, 13 Nov 2022 12:07:38 +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: <y3dp+hintihns...@mkiii.casa>
Content-Type: text/plain; charset=utf-8

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


------------------------------

Subject: Digest Footer

_______________________________________________
Beginners mailing list
Beginners@haskell.org
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners


------------------------------

End of Beginners Digest, Vol 166, Issue 1
*****************************************

Reply via email to