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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 *****************************************