Combining lenses

Question

Using a lens library I can apply a modification function to individual targets, like so:

Prelude Control.Lens> (1, 'a', 2) & _1 %~ (*3)
(3,'a',2)
Prelude Control.Lens> (1, 'a', 2) & _3 %~ (*3)
(1,'a',6)

How can I combine those individual lenses (_1 and _3) to be able to perform this update to both of the targets at once? I expect something in the spirit of the following:

Prelude Control.Lens> (1, 'a', 2) & ??? %~ (*3)
(3,'a',6)

Answer 1

There is a variation on what you are asking for which is more general:

(/\)
    :: (Functor f)
    => ((a -> (a, a)) -> (c -> (a, c)))
    -- ^ Lens' c a
    -> ((b -> (b, b)) -> (c -> (b, c)))
    -- ^ Lens' c b
    -> (((a, b) -> f (a, b)) -> (c -> f c))
    -- ^ Lens' c (a, b)
(lens1 /\ lens2) f c0 =
    let (a, _) = lens1 (\a_ -> (a_, a_)) c0
        (b, _) = lens2 (\b_ -> (b_, b_)) c0
        fab = f (a, b)
    in fmap (\(a, b) ->
            let (_, c1) = lens1 (\a_ -> (a_, a)) c0
                (_, c2) = lens2 (\b_ -> (b_, b)) c1
            in c2
            ) fab

infixl 7 /\

Just focus on the type signature with lens type synonyms:

Lens' c a -> Lens' c b -> Lens' c (a, b)

It takes two lenses and combines them into a lens to a pair of fields. This is slightly more general and works for combining lenses that point to fields of different types. However, then you'd have to mutate the two fields separately.

I just wanted to throw this solution out there in case people were looking for something like this.

Answer 2

Using untainted from the Settable type class in Control.Lens.Internal.Setter, it is possible to combine two setters, but the result will also only be a setter and not a getter.

import Control.Lens.Internal.Setter

-- (&&&) is already taken by Control.Arrow
(~&~) :: (Settable f) => (c -> d -> f a) -> (c -> a -> t) -> c -> d -> t
(~&~) a b f = b f . untainted . a f

You can test this:

>>> import Control.Lens
>>> (1, 'a', 2) & (_1 ~&~ _3) %~ (*3)
(3,'a',6)

EDIT

You don't actually need to use internal functions. You can use the fact that Mutator is a monad:

{-# LANGUAGE NoMonomorphismRestriction #-}

import Control.Monad
import Control.Applicative

(~&~) = liftA2 (>=>)

-- This works too, and is maybe easier to understand: 
(~&~) a b f x = a f x >>= b f