On Sat, Jun 16, 2012 at 3:47 AM, Dan Burton <danburton.em...@gmail.com> wrote:
>
> Convenience aside, doesn't the functor instance conceptually violate some
> sort of law?
>
> fmap (const 1) someSet
>
> The entire shape of the set changes.
>
> fmap (g . h) = fmap g . fmap h
>
> This law wouldn't hold given the following contrived ord instance
>
> data Foo = Foo { a, b :: Int }
> instance Ord Foo where
> compare = compare `on` a
>
> Given functions
>
> h foo = foo { a = 1 }
> g foo = foo { a = b foo }
>
> Does this library address this? If so, how? If not, then you'd best note it
> in the docs.
Your hypothesis is false. You should at least try out your example.
It's easy to show that (fmap g . fmap h) x is true for all x (at least
ignoring potential issues with strictness.) The thing to note is that
fmap h x is not a set, it is an expression tree (which is only
observable via run.) The law that ends up failing is toList .
fromList /= id, i.e. fmap g . toList . fromList . fmap h /= fmap g .
fmap h
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