Ah, yeah, I see that now, in the "natural" comment. In that case I'm not
sure if I prefer the version with FlexibleContexts or QuantifiedConstraints.
On Fri, Nov 30, 2018, 5:18 PM David Feuer I'm suggesting that the lifting instance is the *wrong one*. Lifting is
Ap's job. Compose is just for composing things. On Fri, Nov 30, 2018, 5:14 PM Daniel Cartwright Yeah, the instance being defined with Ap is certainly a good way to go.
We can use DerivingVia and Data.Monoid.Ap. On Fri, Nov 30, 2018, 4:55 PM David Feuer The alternate one can be written much more simply with
FlexibleContexts instead of QuantifiedConstraints: instance Semigroup (f (g a)) => Semigroup (Compose f g a) where
Compose x <> Compose y = Compose (x <> y)
instance Monoid (f (g a)) => Monoid (Compose f g a) where
mempty = Compose mempty That strikes me as the most natural way to do it. The lifting
instances are more naturally done using this oft-proposed type: newtype Ap f a = Ap (f a)
instance (Applicative f, Semigroup a) => Semigroup (Ap f a) where
Ap x <> Ap y = Ap (liftA2 (<>) x y)
instance (Applicative f, Monoid a) => Monoid (Ap f a) where
mempty = pure mempty Now Ap (Compose f g) a will do the lifting you want. On Fri, Nov 30, 2018 at 4:13 PM Daniel Cartwright Data.Functor.Compose: newtype Compose f g a = Compose { getCompose :: f (g a) } there's clear-cut, more traditional instances if `f` and `g` are Applicative: instance (Applicative f, Applicative g, Semigroup a) => Semigroup (<>) = liftA2 (<>) instance (Applicative f, Applicative g, Monoid a) => Monoid (Compose f
g a) where
mempty = pure mempty There's an alternative with `QuantifiedConstraints`, but it's arguable (Compose f g a) where
that this is desirable: instance (forall x. Semigroup x => Semigroup (f x), forall x. Compose x <> Compose y = Compose (x <> y) Both to seem to fit the commonplace spirit of lifting monoids up Semigroup x => Semigroup (g x), Semigroup a) => Semigroup (Compose f g a)
where
through applicative contexts. _______________________________________________
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