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Think `data Foo a = Foo (a -> Bool)`, pardon my Haskell. A function `map :: (a -> b) -> Foo a -> Foo b` is impossible, however `contramap :: (a -> b) -> Foo b -
by antisemiotic 7y ago
Think `data Foo a = Foo (a -> Bool)`, pardon my Haskell. A function `map :: (a -> b) -> Foo a -> Foo b` is impossible, however `contramap :: (a -> b) -> Foo b -> Foo a` is fine (just pre-compose the given function with the stored function).
Even worse with `data Bar a = Bar (a -> a)`.
- TheAsprngHacker 7y agoThanks. I know what contravariant functors are, but I haven't used them, so I didn't realize that was what the parent commenter was asking about. You're right, my claim that every type 'a t has a corresponding (covariant) functor is incorrect, and I should either take that out or mention contravariant functors.
- moomin 7y agoIt’s even worse than that. Consider data Foo a = Foo (a -> a) This admits neither Functor nor Contravariant. Sadly all you can say is “If you can map, it’s a functor.” I always end up finding out more about any subject I actually publish a post about when people read it...
- TheAsprngHacker 7y agoThe other person also mentioned this. It's called invariance. There's also a fourth variance: https://www.benjamin.pizza/posts/2019-01-11-the-fourth-type-of-variance.html https://www.benjamin.pizza/posts/2019-01-11-the-fourth-type-...
- TheAsprngHacker 7y agoThank you for pointing out contravariant functors; I have fixed that sentence in my post and credited you.
- moomin 7y agoOh that’s much more extensive. Like the link to Julie’s work.
- lalaithion 7y agoI believe that you can say "As long as `a` only appears as the last parameter in a higher kinded type, or not as the parameter of any higher kinded type, then you can define a functor over it."