[Haskell-cafe] Representation of lenses
dct25-561bs at mythic-beasts.com
Tue Jan 27 17:50:44 UTC 2015
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
<tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk> wrote:
> 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.
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