[Haskell-cafe] Applicative banana brackets

Kim-Ee Yeoh ky3 at atamo.com
Mon Dec 14 15:41:43 UTC 2015

On Mon, Dec 14, 2015 at 10:10 PM, Erik Hesselink <hesselink at gmail.com>

Every Arrow is a Functor through:
>     fmapA :: Arrow arr => (a -> b) -> arr i a -> arr i b
>     fmapA f a = arr f . a
> Right?

That's one of the missing holes in Martin's claim.

In cases like this, it would help to avoid any risk that the usual abuse of
language brings. So an arrow is not a functor but it does give rise to one.
More precisely, there would be an instance Arrow a => Functor (a b).

-- Kim-Ee
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