On Tue, Jan 27, 2015 at 05:18:41PM +0000, David Turner wrote:
I believe that the types (a -> s, s -> a -> a) and (forall f. Functor f => (s -> f s) -> a -> f a) are isomorphic. Is that right?
Yes (with the caveat that I don't actually know how to prove this).
It's also possible that you can't get the full generality of (forall f. (s -> f t) -> a -> f b) lenses with getter/setter pairs
No, they are equivalent.
So, my question is: what is the advantage of representing lenses in the way that they are?
The advantage that's most obvious to me is polymorphism. You can use a lens forall f. Functor f => (s -> f s) -> a -> f a where the callee expects a traversal forall f. Applicative f => (s -> f s) -> a -> f a This is very convenient in practice. Perhaps there are other practical advantages that someone else can explain. Tom