Re: [Haskell-beginners] How to create a monad in GHC 7.10 or newer

13 Nov 2022


      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  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
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