For the record, I think the isomorphism uses the Const s functor to
recover the getter and the Identity functor to recover the setter.
It's something like this:
Given (get, set), the corresponding lens is \f x. fmap (flip set x) (f (get x))
Given a lens ell the getter/setter pair is (runConst . ell Const, \v
-> runIdentity . ell (Identity . const v))
Notice that Const s is not an Applicative (e.g. pure :: () -> Const
Void () doesn't exist) which is why (forall f. Applicative f => (s ->
f s) -> a -> f a) proscribes getting your hands on the getter.
On 27 January 2015 at 17:25, Tom Ellis
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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe