Jan Hrcek edited this page Jun 3, 2018 · 21 revisions

In order to derive the type signatures used in this package and to motivate generalizing these combinators to work on a much wider menagerie of types, we'll take a bit of an excursion through other types and typeclasses you may already be familiar with.

From here, we'll assume that you are acquainted with the types for Functor, Foldable and Traversable and can at least hand-wave the laws for them (see Typeclassopedia).

The power of (.)

There is a common folklore pattern for composing (.) in Haskell, to compose with a function that takes more and more arguments (see this gist if it isn't obvious).

(.)         :: (a -> b) -> (c -> a)           -> c           -> b
(.).(.)     :: (a -> b) -> (c -> d -> a)      -> c -> d      -> b
(.).(.).(.) :: (a -> b) -> (c -> d -> e -> a) -> c -> d -> e -> b

It generalizes to fmap from Functor:

fmap           :: Functor f
               => (a -> b) -> f a         -> f b
fmap.fmap      :: (Functor f, Functor g)
               => (a -> b) -> f (g a)     -> f (g b)
fmap.fmap.fmap :: (Functor f, Functor g, Functor h)
               => (a -> b) -> f (g (h a)) -> f (g (h b))

It also generalizes to foldMap from Foldable:

foldMap                 :: (Foldable f, Monoid m)
                        => (a -> m) -> f a         -> m
foldMap.foldMap         :: (Foldable f, Foldable g, Monoid m)
                        => (a -> m) -> f (g a)     -> m
foldMap.foldMap.foldMap :: (Foldable f, Foldable g, Foldable h, Monoid m)
                        => (a -> m) -> f (g (h a)) -> m

And finally (for now) it generalizes to traverse from Traversable:

traverse                   :: (Traversable f, Applicative m)
                           => (a -> m b) -> f a         -> m (f b)
traverse.traverse          :: (Traversable f, Traversable g, Applicative m)
                           => (a -> m b) -> f (g a)     -> m (f (g b))
traverse.traverse.traverse :: (Traversable f, Traversable g, Traversable h, Applicative m)
                           => (a -> m b) -> f (g (h a)) -> m (f (g (h b)))

These functions are similar, but slightly different in signature, and provide very different capabilities.

We gain power as we upgrade from Functor or Foldable to Traversable, as evidenced by the fact that we can define fmap and foldMap given traverse. This is done for us directly by Data.Traversable, as follows:

-- | This function may be used as a value for `fmap` in a `Functor`
--   instance, provided that 'traverse' is defined. (Using
--   `fmapDefault` with a `Traversable` instance defined only by
--   'sequenceA' will result in infinite recursion.)
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f = runIdentity . traverse (Identity . f)
-- | This function may be used as a value for `Data.Foldable.foldMap`
-- in a `Foldable` instance.
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f = getConst . traverse (Const . f)

If we rip traverse out of their definitions and pass it in as an argument, we can find out exactly what properties are needed of traverse to make those definitions typecheck. We'll deal with fmapDefault first.


ghci> let over l f = runIdentity . l (Identity . f)
ghci> :t over
over :: ((a -> Identity b) -> s -> Identity t) -> (a -> b) -> s -> t

So lets define a type alias for that:

type Setter s t a b = (a -> Identity b) -> s -> Identity t

which lets us write this as:

over :: Setter s t a b -> (a -> b) -> s -> t
-- or
over :: ((a -> Identity b) -> s -> Identity t)
     ->  (a -> b)          -> s -> t
over l f = runIdentity . l (Identity . f)

Setter Laws

We can write an inverse of over fairly mechanically:

sets :: ((a -> b) -> s -> t) -> Setter s t a b
-- or
sets :: ((a -> b)          -> s -> t)
     ->  (a -> Identity b) -> s -> Identity t
sets l f = Identity . l (runIdentity . f)

It is trivial to verify that

sets . over = id
over . sets = id

but if we want the Setter to act like a Functor, the argument the user supplies to sets should satisfy a version of the Functor laws. In particular, we'd like:

over l id = id
over l f . over l g = over l (f . g)

so the function m you supply to sets must satisfy

m id = id
m f . m g = m (f . g)

(Note: Unlike with fmap we have to actually check both laws, because parametricity doesn't help us derive the second for free from the first.)

So, what satisfies the functor laws out of the box? fmap!

mapped :: Functor f => Setter (f a) (f b) a b
-- or
mapped :: Functor f => (a -> Identity b) -> f a -> Identity (f b)
mapped = sets fmap

It is trivial then to check that

fmap = over mapped

Now we can write a number of combinators that are parameterized on the Functor-like construction they work over.

Lets consider the limited form of mapping allowed on the arrays provided by Data.Array.IArray:

amap :: (IArray a e', IArray a e, Ix i) => (e' -> e) -> a i e' -> a i e

Given this, we can derive:

amapped :: (IArray a c, IArray a d, Ix i) => Setter (a i c) (a i d) c d
-- or
amapped :: (IArray a c, IArray a d, Ix i)
        => (c -> Identity d) -> a i c -> Identity (a i d)
amapped = sets amap

Then we can use

over amapped :: (IArray a c, IArray a d, Ix i) => (c -> d) -> a i c -> a i d

instead of amap.

We can also pass in the type of map provided by, say,

tmapped :: Setter Text Text Char Char
-- or
tmapped :: (Char -> Identity Char) -> Text -> Identity Text
tmapped = sets

And it follows that:

over tmapped :: (Char -> Char) -> Text -> Text

We haven't gained much power over just passing in the functions amap or directly, yet, but we have gained two things:

  • The composition of two setters, such as:

    mapped.mapped :: (Functor f, Functor g) => (a -> Identity b) -> f (g a) -> Identity (f (g b))

    is still a valid Setter!

    mapped.mapped :: (Functor f, Functor g) => Setter (f (g a)) (f (g b)) a b

    So you can use over (mapped.mapped) to recover the original fmap.fmap above. This lets you get away without using Compose to manually bolt functors to meet the shape requirements.

  • Another thing that we have won is that if we have a Traversable container, we can pass its traverse in to over instead of a Setter for the container.

Setters form a category, using (.) and id for composition and identity, but you can use the existing (.) and id from the Prelude for them (though they compose backwards).

However, to gain that power we traded in other functionality. Knowing f is a Functor lets us instantiate the type arguments a and b to anything we want, over and over again, we also need to manually check any of the formerly free theorems we want to use with our Setter.

Many combinators for these are provided in Control.Lens.Setter.

As an aside, we can now define the (.~) (and set) combinators we used during the introduction:

(.~), set :: Setter s t a b -> b -> s -> t
l .~ d = runIdentity . l (Identity . const d)
set = (.~)

In a few moments we'll see how they can be applied to a Lens, but first:


Now lets apply the same treatment to the other default definition supplied by Data.Traversable:

foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f = getConst . traverse (Const . f)

If we plug in an argument for traverse and rip off the type signature, we get

ghci> let foldMapOf l f = getConst . l (Const . f)
ghci> :t foldMapOf
foldMapOf :: ((a -> Const r b) -> s -> Const r t) -> (a -> r) -> s -> r

The second argument to Const is polymorphic in each case, but to avoid dangling type variables we'll eliminate them by making them match.

Writing it out, and making up a type alias:

type Getting r s a = (a -> Const r a) -> s -> Const r s

we can make the slightly nicer looking type

foldMapOf :: Getting r s a -> (a -> r) -> s -> r
-- or
foldMapOf :: ((a -> Const r a) -> s -> Const r s)
          ->  (a -> r)         -> s -> r
foldMapOf l f = getConst . l (Const . f)

It follows by substitution that

foldMapDefault = foldMapOf traverse

we could define an inverse of foldMapOf as we did with over, above, etc.

folds :: ((a -> r) -> s -> r) -> Getting r s a
-- or
folds :: ((a -> r)         -> s -> r)
      ->  (a -> Const r a) -> s -> Const r s
folds l f = Const . l (getConst . f)

If follows very easily that

folds . foldMapOf = id
foldMapOf . folds = id

and so we can see that we can pass something other than traverse to foldMapOf:

type Fold s t a b = forall m. Monoid m => (a -> Const m b) -> s -> Const m t
folded :: Foldable f => Fold (f a) (f a) a a
-- or
folded :: (Foldable f, Monoid m) => (a -> Const m a) -> f a -> Const m (f a)
folded = folds foldMap

And we can mechanically verify that foldMap = foldMapOf folded by:

foldMap = (foldMapOf . folds) foldMap = foldMapOf (folds foldMap) = foldMapOf folded

There are no laws for Foldable that do not follow directly from the types, and the same holds for a Fold.

We can define all of the combinators in Data.Foldable in terms of foldMap, so we in turn define them in terms of an arbitrary Fold in Control.Lens.Fold.

(In the actual implementation the type of Fold is changed to use a typeclass constraint rather than an explicit Const m to yield nicer error messages when you attempt to use a Setter as a Fold and to permit the use of certain Applicative transformers as Monoid transformers.)

As with Setter, the composition of two folds using (.) is a valid Fold, and id is the identity fold that returns the container itself as its only result.


Given the signatures of Fold and Setter, we can derive something that can be used as both -- after all traverse from Traversal served this function originally!

Given the Monoid m, Const m forms an Applicative, and Identity is also Applicative.

So substituting back in to the definitions above, we find:

type SimpleTraversal s a = forall f. Applicative f => (a -> f a) -> s -> f s

But this is weaker than what we started with, since

traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)


type Traversal s t a b = forall f. Applicative f => (a -> f b) -> s -> f t

we get something that subsumes

traverse :: Traversable t => Traversal (t a) (t b) a b
-- or
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)

We'll still be dealing with those 2-argument traversals a lot, along with their Setter equivalents, so we'll define

type Simple f a b = f a a b b

(Note: 'Simple' 'Traversal' partially applies a type synonym. This is only legal with the LiberalTypeSynonyms extension turned on, but GHC allows it so long as the type synonym contains a higher-rank type, even without LiberalTypeSynonyms.)

But, what happened?

When I picked the type for Getting, I only used two type arguments.

type Getting r s a = (a -> Const r a) -> s -> Const r s

This hints that for something to be a valid Traversal it should be possible to choose s ~ t, and a ~ b and get a meaningful traversal. In fact the Traversable laws, which we still want to have hold for a Traversal, tell us that:

l pure = pure
Compose . fmap (l f) . l g = l (Compose . fmap f . g)

And the first of those laws requires s ~ t, a ~ b to be a possible choice of the type arguments to your Traversal. We implement this polymorphic overloading of traversals in a fairly ad hoc way, by just making the user provide the "family structure" for us.

Given that our Traversal satisfies the Traversable laws, the laws for Setter immediately follow, and Fold had no extra laws to check.

So what else can we build a Traversal for?

We can traverse both elements in a tuple:

both :: Traversal (a,a) (b,b) a b
-- or
both :: Applicative f => (a -> f b) -> (a,a) -> f (b,b)
both f (a,b) = (,) <$> f a <*> f b

The left side of an Either:

traverseLeft :: Traversal (Either a c) (Either b c) a b
-- or
traverseLeft :: Applicative f => (a -> f b) -> Either a c -> f (Either b c)
traverseLeft f (Left a) = Left <$> f a
traverseLeft f (Right b) = pure $ Right b

And we can use a Traversal to update element (if it exists) at a given position in a Map, etc.

We can compose two traversals with each other using (.) as we did at the very start, but now the composition forms a valid Traversal.

traverse.traverse :: (Traversable f, Traversable g) => Traversal (f (g a)) (f (g b)) a b

but we can also compose them with a Setter or Fold, yielding a Setter or Fold in turn.

Unlike with mapped and folded, we will often want to use the Traversal directly as a combinator.

Moreover, all the intuition you have for how to write Traversable instances transfers immediately to how to write a Traversal.

There are a number of combinators for working with traversals in Control.Lens.Traversal

We're almost ready for lenses, but first we have one more diversion.


If we convert a function from (a -> c) to continuation passing style, we get

cps :: (a -> c) -> (c -> r) -> a -> r
cps f g = g . f

That is to say, rather than return a result c, it takes a function from (c -> r) and calls it with the answer instead.

If we have a CPS'd function that is polymorphic in its return type, we can get the original function back out:

uncps :: (forall r. (c -> r) -> a -> r) -> a -> c
uncps f = f id

Finally, we can prove a number of properties:

uncps . cps = id
cps . uncps = id
cps id = id
cps f . cps g = cps (g . f)

Now, we can relax the type of uncps slightly to

uncps' :: ((c -> c) -> a -> c) -> a -> c
uncps' f = f id

but now we no longer know that our function f :: (c -> c) -> a -> c can't be doing something to combine the results of the function you pass it. We lose the cps . uncps = id law and only have:

uncps' . cps  = id

Now, let's compose them like we did for fmap, traverse and foldMap at the start:

If we start with 3 functions:

f :: A -> B
g :: B -> C
h :: C -> D


cps f :: (B -> r) -> (A -> r)
cps g :: (C -> r) -> (B -> r)
cps h :: (D -> r) -> (C -> r)

And when we compose these functions between functions, we obtain:

cps f                 :: (B -> r) -> (A -> r)
cps f . cps g         :: (C -> r) -> (A -> r)
cps f . cps g . cps h :: (D -> r) -> (A -> r)

Earlier we provided a type for consuming a Fold:

type Getting r s a = (a -> Const r a) -> s -> Const r s

What we want (so that uncps can work) is something completely polymorphic in r.

type Getter s a = forall r. (a -> Const r a) -> s -> Const r s

Along the way, we get an interesting result: a Getter is just a Fold that doesn't use the Monoid! Recall:

type Fold s a = forall r. Monoid r => (a -> Const r a) -> s -> Const r s

We can go back and define (^.) now, and empower it to consume either a Fold or Getter or Traversal.

(^.) :: s -> Getting a s a -> a
s ^. l = getConst (l Const s)

Remember, we can consume a Traversal because every Traversal is a valid Fold, just like every Getter is a valid Fold.

Also note that (^.) doesn't require anything of a!

When it gets applied, the argument l will demand the properties of a that it needs:

For instance when we apply (^.) to a Fold, it will demand a Monoid instance for a:

(^.folded) :: (Foldable f, Monoid m) => f m -> m

Also, since, a Monoid m is needed to satisfy the Applicative for Const m,

(^.traverse) :: (Traversable t, Monoid m) => t m -> m

But we can use (^.) to access a Getter, without any restrictions!

There are a number of combinators for working with getters in Control.Lens.Getter.

We can use a Getter as a Fold, but it is not a valid Traversal or Setter, nor is a Traversal or Setter a Getter.

With all of that we're finally ready to define


A Lens is a Traversal that can also be used as a Getter. This means it can be used as a Setter and a Fold as well.

Recall that a Getter was a Fold that can't use the Monoid.

Without the Monoid, all that Const and Identity have in common is that each is a Functor.

type Lens s t a b = forall f. Functor f => (a -> f b) -> s -> f t

We inherit the Traversal laws, so we know for a Lens l

l pure = pure
Compose . fmap (l f) . l g = l (Compose . fmap f . g)

and we also know that a Lens s t a b can be used as a function from (s -> a).

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