Lenses are Static Selectors

So I don’t really know what KVC is, or much about performSelector functions. This blogpost, from Brent Simmons, let me know a little bit about why I would want to use them.

It centred around removing code repetition of this type:

if localObject.foo != serverObject.foo {
  localObject.foo = serverObject.foo

if localObject.bar != serverObject.bar {
  localObject.bar = serverObject.bar // There was an (intentional)
}                                    // bug here in the original post

To clean up the code, Brent used selector methods. At first, I was a little uncomfortable with the solution. As far as I could tell, the basis of a lot of this machinery used functions with types like this:

func get(fromSelector: String) -> AnyObject?
func set(forSelector: String) -> ()

Which seems to be extremely dynamic. Stringly-typed and all that. Except that there are two different things going on here. One is the dynamic stuff; the ability to get rid of types when you need to. The other, though, has nothing to do with types. The other idea is being able to pass around something which can access the property (or method) of an object.

Let’s look at the code that was being repeated:

if localObject.foo != serverObject.foo {
  localObject.foo = serverObject.foo

if localObject.bar != serverObject.bar {
  localObject.bar = serverObject.bar

The logical, obvious thing to do here is try refactor out the common elements. In fact, the only things that differ between the two actions above are the foo and bar. It would be great to be able to write a function like this:

func checkThenUpdate(selector) {
  if localObject.selector != serverObject.selector {
    localObject.selector = serverObject.selector

And then maybe a single line like this:

[foo, bar, baz].forEach(checkThenUpdate)

That’s pretty obviously better. It’s just good programming: when faced with repetition, find the repeated part, and abstract it out. Is it more dynamic than the repetition, though? I don’t think so. All you have to figure out is an appropriate type for the selector, and you can keep all of your static checking. To me, it seems a lot like a lens:

struct Lens<Whole, Part> {
  let get: Whole -> Part
  let set: (Whole, Part) -> Whole

(This is a lens similar to the ones used in the data-lens library, in contrast to van Laarhoven lenses, or LensFamilies. LensFamilies are used in the lens package, and they allow you to change the type of the Part. They’re also just normal functions, rather than a separate type, so you can manipulate them in a pretty standard way. Swift’s type system isn’t able to model those lenses, though, unfortunately.)

It has two things: a getter and a setter. The getter is pretty obvious: it takes the object, and returns the property. The setter is a little more confusing. It’s taking an object, and the new property you want to stick in to the object, and returns the object with that property updated.

For instance, if we were to make a Person:

struct LocalPerson {
  var age: Int
  var name: String

We could then have a lens for the name field like this:

let localName: Lens<LocalPerson,String> = Lens(
  get: { p in p.name },
  set: { (oldPerson,newName) in
    var newPerson = oldPerson
    newPerson.name = newName
    return newPerson

And you’d use it like this:

let caoimhe = LocalPerson(age: 46, name: "caoimhe")
localName.get(caoimhe) // 46
localName.set(caoimhe, "breifne") // LocalPerson(age: 46, name: "breifne")

Straight away, we’re able to do (something) like the checkThenUpdate function:

func checkThenUpdate
  <A: Equatable>
  (localLens: Lens<LocalPerson,A>, serverLens: Lens<ServerPerson,A>) {
  let serverProp = serverLens.get(serverObject)
  if localLens.get(localObject) != serverProp {
    localObject = localLens.set(localObject,serverProp)

And it could be called pretty tersely:

checkThenUpdate(localName, serverLens: serverName)

The biggest problem with this approach, obviously, is the boilerplate. In Haskell, that’s solved with Template Haskell, so the lens code is generated for you. (I’d love to see something like that in Swift)

There’s a protocol-oriented spin on lenses, also. One of the variants on lenses in Haskell are called "classy-lenses". That’s where, instead of just generating a lens with the same name as the field it looks into, you generate a typeclass (protocol) for anything with that lens. In Swift, it might work something like this:

struct Place {
  var name: String

// Instead of just having a lens for the name field, have a whole protocol
// for things with a name field:

protocol HasName {
  associatedtype Name
  static var name: Lens<Self,Name> { get }
  var name: Name { get set }

// Because the mutable property is included in the protocol, you can rely on
// it in extensions:

extension HasName {
  static var name: Lens<Self,Name> {
    return Lens(
      get: {$0.name},
      set: { (w,p) in 
        var n = w
        n.name = p
        return n
  var name: Name {
    get { return Self.name.get(self) }
    set { self = Self.name.set(self,newValue) }

// This way, you can provide either the lens or the property, and you get the
// other for free.

extension Place: HasName {}

// Then, you can rely on that protocol, and all of the types:

func checkEqualOnNames
  <A,B where A: HasName, B: HasName, A.Name: Equatable, A.Name == B.Name>
  (x: A, _ y: B) -> Bool {
    return x.name == y.name

This protocol lets you do a kind of static respondsToSelector, with all of the types intact.

Other people have spoken about the other things you can do with lenses in Swift (Brandon Williams – Lenses in Swift), like composing them together, chaining operations, etc. (One other thing they can emulate is method cascading) Unfortunately, in current Swift, the boilerplate makes all of this a little unpleasant. Still, they’re an interesting idea, and they show how a good type system needn’t always get in the way.

Folding Two Things at Once

There’s a whole family of Haskell brainteasers surrounding one function: foldr. The general idea is to convert some function on lists which uses recursion into one that uses foldr. map, for instance:

map :: (a -> b) -> [a] -> [b]
map f = foldr (\e a -> f e : a) []

Some can get a little trickier. dropWhile, for instance. (See here and here for interesting articles on that one in particular.)

dropWhile :: (a -> Bool) -> [a] -> [a]
dropWhile p = fst . foldr f ([],[]) where
  f e ~(xs,ys) = (if p e then xs else zs, zs) where zs = e : ys


One function which was a little harder to convert than it first seemed was zip.

Here’s the first (non) solution:

zip :: [a] -> [b] -> [(a,b)]
zip = foldr f (const []) where
  f x xs (y:ys) = (x,y) : xs ys
  f _ _  [] = []

The problem with the above isn’t that it doesn’t work: it does. The problem is that it’s not really using foldr. It’s only using it on the first list: there’s still a manual uncons being performed on the second. Ideally, I would want the function to look something like this:

zip :: [a] -> [b] -> [(a,b)]
zip xs ys = foldr f (\_ _ -> []) xs (foldr g (const []) ys)

The best solution I found online only dealt with Folds, not Foldables. You can read it here.

Recursive Types

Reworking the solution online for Foldables, the initial intuition is to have the foldr on the ys produce a function which takes an element of the xs, and returns a function which takes an element of the xs, and so on, finally returning the created list. The problem with that approach is the types involved:

zip :: [a] -> [b] -> [(a,b)]
zip xs = foldr f (const []) xs . foldr g (\_ _ -> []) where
  g e2 r2 e1 r1 = (e1,e2) : (r1 r2)
  f e r x = x e r

You get the error: Occurs check: cannot construct the infinite type: t0 ~ a -> (t0 -> [(a, b)]) -> [(a, b)]. Haskell’s typechecker doesn’t allow for infinitely recursive types.

You’ll be familiar with this problem if you’ve ever tried to encode the Y-combinator, or if you’ve fiddled around with the recursion-schemes package. You might also be familiar with the solution: a newtype, encapsulating the recursion. In this case, the newtype looks very similar to the signature for foldr:

newtype RecFold a b = 
  RecFold { runRecFold :: a -> (RecFold a b -> b) -> b }

Now you can insert and remove the RecFold wrapper, helping the typechecker to understand the recursive types as it goes:

zip :: [a] -> [b] -> [(a,b)]
zip xs =
  foldr f (const []) xs . RecFold . foldr g (\_ _ -> []) where
    g e2 r2 e1 r1 = (e1,e2) : (r1 (RecFold r2))
    f e r x = runRecFold x e r

As an aside, the performance characteristics of the newtype wrapper are totally opaque to me. There may be significant improvements by using coerce from Data.Coerce, but I haven’t looked into it.

Generalised Zips

The immediate temptation from the function above is to generalise it. First to zipWith, obviously:

zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith c xs =
  foldr f (const []) xs . RecFold . foldr g (\_ _ -> []) where
    g e2 r2 e1 r1 = c e1 e2 : (r1 (RecFold r2))
    f e r x = runRecFold x e r

What’s maybe a little more interesting, though, would be a foldr on two lists. Something which folds through both at once, using a supplied combining function:

foldr2 :: (Foldable f, Foldable g)
       => (a -> b -> c -> c)
       -> c -> f a -> g b -> c
foldr2 c i xs =
  foldr f (const i) xs . RecFold . foldr g (\_ _ -> i) where
    g e2 r2 e1 r1 = c e1 e2 (r1 (RecFold r2))
    f e r x = runRecFold x e r

Of course, once you can do two, you can do three:

foldr3 :: (Foldable f, Foldable g, Foldable h)
       => (a -> b -> c -> d -> d)
       -> d -> f a -> g b -> h c -> d
foldr3 c i xs ys =
  foldr f (const i) xs . RecFold . foldr2 g (\_ _ -> i) ys where
    g e2 e3 r2 e1 r1 = c e1 e2 e3 (r1 (RecFold r2))
    f e r x = runRecFold x e r

And so on.

There’s the added benefit that the above functions work on much more than just lists.


Getting a little formal about the above functions, a fold can be described as a catamorphism. This is a name for a pattern of breaking down some recursive structure. There’s a bunch of them in the recursion-schemes package. The question is, then: can you express the above as a kind of catamorphism? Initially, using the same techniques as before, you can:

newtype RecF f a = RecF { unRecF :: Base f (RecF f a -> a) -> a }

zipo :: (Functor.Foldable f, Functor.Foldable g)
     => (Base f (RecF g c) -> Base g (RecF g c -> c) -> c)
     -> f -> g -> c
zipo alg xs ys = cata (flip unRecF) ys (cata (RecF . alg) xs)

Then, coming full circle, you get a quite nice encoding of zip:

zip :: [a] -> [b] -> [(a,b)]
zip = zipo alg where
  alg Nil _ = []
  alg _ Nil = []
  alg (Cons x xs) (Cons y ys) = (x, y) : ys xs

However, the RecF is a little ugly. In fact, it’s possible to write the above without any recursive types, using the RankNTypes extension. (It’s possible that you could do the same with foldr2 as well, but I haven’t figured it out yet)

You can actually use a newtype that’s provided by the recursion-schemes library as-is. It’s Mu. This is required for an encoding of the Y-combinator. It’s usually presented in this form:

newtype Mu a = Roll { unroll :: Mu a -> a }

However, in the recursion-schemes package, its definition looks like this:

newtype Mu f = Mu (forall a. (f a -> a) -> a)

No recursion! The zipo combinator above can be written using Mu like so:

zipo :: (Functor.Foldable f, Functor.Foldable g)
     => (Base f (Mu (Base g) -> c) -> Base g (Mu (Base g)) -> c)
     -> f -> g -> c
zipo alg xs = cata (\x -> alg x . project) xs . refix

And the new version of zip has a slightly more natural order of arguments:

zip :: [a] -> [b] -> [(a,b)]
zip = zipo alg where
  alg Nil _ = []
  alg _ Nil = []
  alg (Cons x xs) (Cons y ys) = (x,y) : xs ys

Zipping Into

There’s one more issue, though, that’s slightly tangential. A lot of the time, the attraction of rewriting functions using folds and catamorphisms is that the function becomes more general: it no longer is restricted to lists. For zip, however, there’s still a pesky list left in the signature:

zip :: (Foldable f, Foldable g) => f a -> g b -> [(a,b)]

It would be a little nicer to be able to zip through something preserving the structure of one of the things being zipped through. For no reason in particular, let’s assume we’ll preserve the structure of the first argument. The function will have to account for the second argument running out before the first, though. A Maybe can account for that:

zipInto :: (Foldable f, Foldable g) 
        => (a -> Maybe b -> c) 
        -> f a -> g b -> f c

If the second argument runs out, Nothing will be passed to the combining function.

It’s clear that this isn’t a fold over the first argument, it’s a traversal. A first go at the function uses the state monad, but restricts the second argument to a list:

zipInto :: Traversable f => (a -> Maybe b -> c) -> f a -> [b] -> f c
zipInto c xs ys = evalState (traverse f xs) ys where
  f x = do
    h <- gets uncons
    case h of 
      Just (y,t) -> do 
        put t
        pure (c x (Just y))
      Nothing -> pure (c x Nothing)

That code can be cleaned up a little:

zipInto :: Traversable f => (a -> Maybe b -> c) -> f a -> [b] -> f c 
zipInto c = evalState . traverse (state . f . c) where
  f x [] = (x Nothing, [])
  f x (y:ys) = (x (Just y), ys)

But really, the uncons needs to go. Another newtype wrapper is needed, and here’s the end result:

newtype RecAccu a b =
  RecAccu { runRecAccu :: a -> (RecAccu a b, b) }
zipInto :: (Traversable t, Foldable f)
        => (a -> Maybe b -> c) -> t a -> f b -> t c
zipInto f xs =
  snd . flip (mapAccumL runRecAccu) xs . RecAccu . foldr h i where
    i e = (RecAccu i, f e Nothing)
    h e2 a e1 = (RecAccu a, f e1 (Just e2))

A Trie in Haskell

Basic Ops

A Trie is one of those data structures that I find myself writing very early on in almost every language I try to learn. It’s elegant and interesting, and easy enough to implement.

I usually write a version that is a set-like data structure, rather than a mapping type, for simplicity’s sake. It stores sequences, in a prefix-tree structure. It has a map (dictionary) where the keys are the first element of every sequence it stores, and the values are the Tries which store the rest of the sequence. It also has a boolean tag, representing whether or not the current Trie is a Trie on which a sequence ends. Here’s what the type looks like in Haskell:

import qualified Data.Map.Lazy as M

data Trie a = Trie { endHere :: Bool
                   , getTrie :: M.Map a (Trie a)
                   } deriving (Eq)

Now, inserting into the Trie is easy. You just uncons on a list, and insert the head into the map, with the value being the tail inserted into whatever existed at that key before:

empty :: Trie a
empty = Trie False M.empty
insert :: Ord a => [a] -> Trie a -> Trie a
insert [] (Trie _ m)     = Trie True m
insert (x:xs) (Trie e m) = Trie e (M.alter (Just . insert xs . fromMaybe empty) x m)

Searching is simple, also. For the empty list, you just check if the Trie has its endHere tag set to True, otherwise, you uncons, search the map, and query the Trie with the tail if it eas found, or just return False if it was not:

member :: Ord a => [a] -> Trie a -> Bool
member [] (Trie e _)     = e
member (x:xs) (Trie _ m) = fromMaybe False (member xs <$> M.lookup x m)

Here’s my problem. Both of those functions have the same pattern:

f []     = ...
f (x:xs) = ...

Any good Haskeller should be begging for a fold at this stage. But it proved a little trickier than I’d imagined. Take member, for instance. You want to fold over a list, with the base case being the tag on the Trie:

member :: Ord a => [a] -> Trie a -> Bool
member = foldr f base where
  base = ???
  f e a = M.lookup e ???

Where do you get the base case from, though? You have to specify it from the beginning, but the variable you’re looking for is nested deeply into the Trie. How can you look into the Trie, without traversing the list, to find the tag, at the beginning of the function?

That had been my issue for a while. Every time I cam back to writing a Trie, I would see the pattern, try and write insert and member with a fold, and remember again the trouble I had had with it in the past. Recently, though, I saw a different problem, that gave me an idea for a solution.

The Highest Order

“Rewrite dropWhile using foldr

It’s a (semi) well-known puzzle, that’s maybe a little more difficult than it seems at first. Here, for instance, was my first attempt at it:

dropWhile' :: (a -> Bool) -> [a] -> [a]
dropWhile' p = foldr f [] where
  f e a | p e       = a
        | otherwise = e:a

Yeah. That’s filter, not dropWhile:

dropWhile' (<5) [1, 3, 6, 3, 1] -- [6]

Here was my final solution:

dropWhile' :: (a -> Bool) -> [a] -> [a]
dropWhile' p l = drop (foldr f 0 l) l where
  f e a | p e       = a + 1
        | otherwise = 0

After the problem I found this issue of The Monad Reader, which talks about the same problem. In my drop version, I had been counting the number of items to drop as I went, adding one for every element that passed the test. The corresponding version in the article had been building up tail functions, using . to add them together:

dropWhile' :: (a -> Bool) -> [a] -> [a]
dropWhile' p l = (foldr f id l) l where
  f e a | p e       = tail . a
        | otherwise = id

A quick visit to pointfree.io can generate some monadic pointsfree magic:

dropWhile' :: (a -> Bool) -> [a] -> [a]
dropWhile' p = join (foldr f id) where
  f e a | p e       = tail . a
        | otherwise = id

Now, the final version in the article did not use this technique, as it was very inefficient. It used some cleverness beyond the scope of this post. The second-from-last version I quite liked, though:

dropWhile' :: (a -> Bool) -> [a] -> [a]
dropWhile' p l = foldr f l l where
  f e a | p e       = tail a
        | otherwise = l

However, the idea of building up a function in a fold gave me an idea for adapting it to some of the Trie functions.

Folding Inwards

Let’s start with member. It needs to fold over a list, and generate a function which acts on a Trie:

member :: Ord a => [a] -> Trie a -> Bool
member = foldr f base

The base is the function being built up: the final part of the function chain. Each part of the function is generated based on each element of the list, and then chained with the base using .:

member = foldr f base where
  f e a = ??? . a 

The base here is what’s called when the list is empty. Here’s what it looked like in the explicit recursion version:

member [] (Trie e _) = e

We could simplify this by using record syntax, and getTrie:

member [] t = getTrie t

And this has an obvious pointsfree version:

member [] = getTrie

That fits for the base case. It’s just a function:

member = foldr f endHere where
  f e a = ??? . a 

Then, how to combine it. That’s easy enough, actually. It accesses the map, searches it for the key, and calls the accumulating function on it. If it’s not found in the map, just return False. Here it is:

member :: Ord a => [a] -> Trie a -> Bool
member = foldr f endHere where
  f e a = fromMaybe False . fmap a . M.lookup e . getTrie

One of the other standard functions for a Trie is returning the “completions” for a given sequence. It’s a very similar function to member, actually: instead of calling endHere on the final Trie found, though, just return the Trie itself. And the thing to return if any given element of the sequence isn’t found is just an empty Trie:

complete :: Ord a => [a] -> Trie a -> Trie a
complete = foldr f id where
  f e a = fromMaybe empty . fmap a . M.lookup e . getTrie 

In fact, you could abstract out the commonality here:

follow :: Ord a => c -> (Trie a -> c) -> [a] -> Trie a -> c
follow ifMiss onEnd = foldr f onEnd where
  f e a = fromMaybe ifMiss . fmap a . M.lookup e . getTrie 
member :: Ord a => [a] -> Trie a -> Bool
member = follow False endHere

complete :: Ord a => [a] -> Trie a -> Trie a
complete = follow empty id

Folding in and out

insert is another deal entirely. In member, the fold was tunneling into a Trie, applying the accumulator function to successively deeper Tries, and returning a result based on the final Trie. insert needs to do the same tunneling – but the Trie returned needs to be the outer Trie.

It turns out it’s not that difficult. Instead of “building up a function” that is then applied to a Trie, here a function is “sent” into the inner Tries. The cool thing here is that the function being sent hasn’t been generated yet.

Here’s some more illustration of what I mean. Start off with the normal foldr:

insert :: Ord a => [a] -> Trie a -> Trie a
insert = foldr f (\(Trie _ m) -> Trie True m)

With the final function to be applied being one that just flips the endHere tag to True. Then f: this is going to act over the map of the Trie that it’s called on. It’s useful to define a function just for that:

overMap :: Ord b => (M.Map a (Trie a) -> M.Map b (Trie b)) -> Trie a -> Trie b
overMap f (Trie e m) = Trie e (f m)

Then, it will look up the next element of the sequence in the Trie, and apply the accumulating function to it. (if it’s not found it will provide an empty Trie instead) Simple!

insert :: Ord a => [a] -> Trie a -> Trie a
insert = foldr f (\(Trie _ m) -> Trie True m) where
  f e a = overMap (M.alter (Just . a . fromMaybe (Trie False M.empty)) e)

I think this is really cool: with just a foldr, you’re burrowing into a Trie, changing it, and burrowing back out again.


This is always the tricky one with a Trie. You can just follow a given sequence down to its tag, and flip it from on to off. But that doesn’t remove the sequence itself from the Trie. So maybe you just delete the sequence – but that doesn’t work either. How do you know that there are no other sequences stored below the one you were examining?

What you need to do is to send a function into the Trie, and have it report back as to whether or not it stores other sequences below it. So this version of foldr is going to burrow into the Trie, like member; maintain the outer Trie, like insert; but also send messages back up to the outer functions. Cool!

The way to do the “message sending” is with Maybe. If the function you send into the Trie to delete the end of the sequence returns Nothing, then it signifies that you can delete that member. Luckily, the alter function on Data.Map works well with this:

alter :: Ord k => (Maybe a -> Maybe a) -> k -> Map k a -> Map k a

Its first argument is a function which is given the result of looking up its second argument. If the function returns Nothing, that key-value pair in the map is deleted (if it was there). If it returns Just something, though, that key-value pair is added. In the delete function, we can chain the accumulating function with =<<. This will skip the rest of the accumulation if any part of the sequence isn’t found. The actual function we’re chaining on is nilIfEmpty, which checks if a given Trie is empty, and returns Just the Trie if it’s not, or Nothing otherwise.

Here’s the finished version:

delete :: Ord a => [a] -> Trie a -> Trie a
delete = (fromMaybe empty .) . foldr f i where
  i (Trie _ m) | M.null m  = Nothing
               | otherwise = Just (Trie False m)
  f e a = nilIfEmpty . overMap (M.alter (a =<<) e) 
null :: Trie a -> Bool
null (Trie e m) = (not e) && (M.null m)

nilIfEmpty :: Trie a -> Maybe (Trie a)
nilIfEmpty t | null t    = Nothing
             | otherwise = Just t

Folding the Foldable

So how about folding the Trie itself? Same trick: build up a function with a fold. This time, a fold over the map, not a list. And the function being built up is a cons operation. When you hit a True tag, fire off an empty list to the built-up function, allowing it to evaluate:

foldrTrie :: ([a] -> b -> b) -> b -> Trie a -> b
foldrTrie f i (Trie a m) = M.foldrWithKey ff s m where
  s    = if a then f [] i else i
  ff k = flip (foldrTrie $ f . (k :))

Unfortunately, it’s not easy to make the Trie conform to Foldable. It is possible, and it’s what I’m currently trying to figure out, but it’s non-trivial.

Monty Hall

The Monty Hall problem is a great example of how counter-intuitive probability can sometimes be. It goes something like this: say you’re on a gameshow, with the chance to win a car. You’re shown three doors, and the car is behind one, goats behind the other two. You pick a door, say the leftmost, but then the host of the gameshow stops you before it’s opened. He opens one of the two doors you didn’t pick, revealing a goat. He then asks you if you’d like to change your decision. So? Do you?

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In which I Misunderstand Dependent Types

All of this available as a shiny, fancy playground here.

So there was a blog post the other day about dependent types:

Why Dependent Types Matter

Most blog posts that deal with the more mathematical/category theory side of programming go over my head when I read them, only to become suddenly clear a few weeks down the line. I have not reached this such sudden clarity with this post, but I’m starting to see a glimmer.

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