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Applicative implementation of Const Monoid


Applicative implementation of Const Monoid

By : bhushan rajput
Date : November 22 2020, 02:59 PM
To fix this issue ,
After changing the names to the current situation:
code :
(<*>) :: c (m -> b) -> c m -> c b
(<*>) :: Const m (a -> b) -> Const m a -> Const m b


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Applicative Instance for (Monad m, Monoid o) => m o?

Applicative Instance for (Monad m, Monoid o) => m o?


By : A. Gonzalez
Date : March 29 2020, 07:55 AM
Does that help You want the Constant applicative from Data.Functor.Constant in the transformers package, which you can find here.
This Applicative has the following instance:
code :
instance (Monoid a) => Applicative (Constant a) where
    pure _ = Constant mempty
    Constant x <*> Constant y = Constant (x `mappend` y)
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
    pure x = Compose (pure (pure x))
    Compose f <*> Compose x = Compose ((<*>) <$> f <*> x)
How to add applicative effects to a monoid?

How to add applicative effects to a monoid?


By : Yatendra Dohare
Date : March 29 2020, 07:55 AM
wish help you to fix your issue How can I change following code so that each element in Maybe monoid element takes also two arguments (so it would become "double" reader applicative)? , This is how I would do it
code :
g2 x y = g $ ($ y) <$> ($ x) <$> theList
How are monoid and applicative connected?

How are monoid and applicative connected?


By : Christian Jungers
Date : March 29 2020, 07:55 AM
help you fix your problem Remark: I don't own the book (yet), and IIRC, at least one of the authors is active on SO and should be able to answer this question. That being said, the idea behind a monoid (or rather a semigroup) is that you have a way to create another object from two objects in that monoid1:
code :
mappend :: Monoid m => m -> m -> m
amappend :: f (a -> b) -> f a -> f b
amappend :: f          -> f   -> f
fmap     ::   (a -> b) -> f a -> f b
--          ^
--       no f here
($)   ::   (a -> b) ->   a ->   b
(<*>) :: f (a -> b) -> f a -> f b
increase  :: Int -> Int
increase x = x + 1

five :: Int
five = 5

increaseA :: Applicative f => f (Int -> Int)
increaseA = pure increase

fiveA :: Applicative f => f Int
fiveA = pure 5

normalIncrease      = increase   $  five
applicativeIncrease = increaseA <*> fiveA
Monoidal Functor is Applicative but where is the Monoid typeclass in the definition of Applicative?

Monoidal Functor is Applicative but where is the Monoid typeclass in the definition of Applicative?


By : coderax
Date : March 29 2020, 07:55 AM
Any of those help The monoid that's referred to with “monoidal functor” is not a Monoid monoid, i.e. a value-level monoid. It's a type-level monoid instead. Namely, the boring product monoid
code :
type Mempty = ()
type a <> b = (a,b)
class Functor f => Monoidal f where
  pureUnit :: f Mempty
  fzip :: f a -> f b -> f (a<>b)

-- an even more “general nonsense”, equivalent formulation is
-- upure :: Mempty -> f Mempty
-- fzipt :: (f a<>f b) -> f (a<>b)
-- i.e. the functor maps a monoid to a monoid (in this case the same monoid).
-- That's really the mathematical idea behind this all.
class Functor f => Monoidal f where
  pureUnit :: f ()
  fzip :: f a -> f b -> f (a,b)
instance Monoid a => Monoidal ((,) a) where
  pureUnit = (mempty, ())
  fzip (p,a) (q,b) = (p<>q, (a,b))
How can Monoid r => Const r of the optic Fold type be generalized to Contravariant + Applicative?

How can Monoid r => Const r of the optic Fold type be generalized to Contravariant + Applicative?


By : user3078291
Date : March 29 2020, 07:55 AM
this will help The Monoid instance is hidden in the constraint of the Applicative instance for Const r.
Const r is only an instance of Applicative if r is an instance of Monoid:
code :
instance Monoid m => Applicative (Const m) where
   ...
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