Existential Quantification in TypeScript

While working on a TypeScript project, I encountered a scenario that seemed impossible to describe with the language. I needed a function that could take a list of action/handler tuples, [Promise<A>, (a: A) => void][], and apply each handler to the the awaited result of the action. Something like this:

async function applyMany<A>(ahts: [Promise<A>, (a: A) => void][]): Promise<void>

The problem is that A constrains every tuple in the list. applyMany should only care about the compatibility of the action and the handler within each tuple.

The solution is Existential Quantification. It can be used to "hide" types that are not meaningful externally. Unfortunately, TypeScript does not explicitly support existential quantification.1 It is possible however, to "wrap" over the extra type with clever use of closure and type inference.2 Examples of this cleverness were initially difficult for me to decipher, in part, due to my shaky understanding of existential quantification.

While researching the topic, I came across a section in the GHC user guide that briefly discusses existential quantification in the context of Haskell. Coincidentally, the example it presents is isomorphic to my problem. I decided to implement the example in TypeScript as an exercise. Before working through this example, let's cover some fundamentals.

#Generic Functions in TypeScript

function id<A>(a: A): A {
  return a
}

The identity function accepts a value of any type A and returns it unmodified.

id<number>(1) // 1
id<string>('foo') // 'foo'
id<boolean>(true) // true

We know intuitively that manually supplying the types is redundant. The compiler is able to infer A from the argument a:

id(1) // 1
id('foo') // 'foo'
id(true) // true

id can also be implemented as an anonymous function:

const id = <A>(a: A): A => a

Put another way:

const id: <A>(a: A) => A = (a) => a

More clearly:

type F = <A>(a: A) => A

const id: F = (a) => a

Typically when we see a generic type alias, the type variables are on the left side of the = like so:

type F<A> = (a: A) => A

Now, we're in a situation where A must always be supplied. It is no longer possible for the compiler to infer the type from the argument. As a result, id must be implemented for every type:

const idNumber: F<number> = (a) => a
const idString: F<string> = (a) => a
const idBoolean: F<boolean> = (a) => a
// ...

This mirrors the original problem:

type AHT<A> = [Promise<A>, (a: A) => void]

const ahtNumber: AHT<number> = [Promise.resolve(0), (a) => console.log(a + 1)]
const ahtString: AHT<string> = [Promise.resolve('foo'), (a) => console.log(a.length)]
const ahtBool: AHT<boolean> = [Promise.resolve(true), (a) => console.log(!a)]
// ...

<A> must move to the right of the = so that an AHT is constructible for all types A. From what we've seen, functions provide a way to perform this transformation.

#A Simple Existential Type

Let's come back to the Haskell example.

data Foo = forall a. MkFoo a (a -> Bool)
         | Nil

Foo has two data constructors: MkFoo, and Nil.

MkFoo :: forall a. a -> (a -> Bool) -> Foo
Nil   :: Foo

MkFoo is clearly the function of interest. forall a. "existentializes" the type variable a removing it from the left side of the =‌. The use of forall to existentially quantify a type variable is the source of much confusion. To avoid introducing an additional keyword exists, the implementers leveraged a recontextualisation of De Morgan's laws within type theory that describes an isomorphism between exists and forall.3 Barring all that ivory tower stuff, I just think of it like this:

MkFoo constructs a Foo forall types a.

Notice that this is a restatement of what was said about the AHT type in the previous section.

With the a extistentialized, a heterogeneous list of Foos can be created.

[MkFoo 3 even, MkFoo 'c' isUpper, Nil] :: [Foo]

In the first Foo, A is Int. In the second, A is Char. How can this be achieved in TypeScript? Let's start by defining Foo.

type Foo = MkFoo | null

We don't need to define Nil because TypeScript (unfortunately) already has a null type. Next we need to define MkFoo. Starting from the bottom...

type MkFoo_<A> = [A, (a: A) => boolean]

MkFoo_ establishes the relationship between the value and the handler. Next, we need to move the <A> to the other side of the = by wrapping MkFoo in a function.

type MkFoo = <R>(run: <A>(_: MkFoo_<A>) => R) => R

Well, there's actually two functions. Let's analyze this layer by layer.

<A>(_: MkFoo_<A>) => R

The inner function infers A from the MkFoo_ argument. As it is a function, it must return something. An additional type variable R is used to avoid constraining what can be returned. With only this layer, R would appear on the left of the = so an additionally layer is necessary.

<R>(run: <A>(_: MkFoo_<A>) => R) => R

This function is able to infer R from the inner layer run. You may recognize the "shape" ((_ -> r) -> r) which is common in continuation passing style programming. This blog post explains existential quantification in TypeScript through the lens of CPS.

Now we need a constructor for MkFoo:

const mkFoo = <A>(mkFoo_: MkFoo_<A>): MkFoo => (run) => run(mkFoo_)

This wraps a MkFoo_ in the layers just described. By accepting a run function, a MkFoo provides the ability for an external entity to "reach in" and safely interact with the hidden MkFoo_. It's a bit like the What's in the Box? challenge.

Finally, we can create the list of Foos:

const foos = [
  mkFoo([3, (a) => a % 2 === 0]),
  mkFoo(['c', (a) => a === a.toUpperCase()]),
  null
]

Notice that we don't have to explicitly fill in any type variables. In the first Foo, A is inferred to be number. In the second, A is inferred to be string. A is now "closed over" but R still needs filling.

const f = (foo: Foo): boolean => {
  if (foo === null) return false
  return foo(([v, h]) => h(v))
}

foos.map(f) // [false, false, false]

Because the return type of f is boolean, R is filled in with boolean. The run function ([v, h]) => h(v), specified to return R, must also return boolean. h(v) returns boolean so the constraint is satisfied. The relationship between v and h is already encoded within foo so they're left unspecified.

We've shown that existential quantification is possible in TypeScript. You can play around with this example here.

After implementing Foo, I was able to easily transform it into a solution for my initial problem. A new problem arose. I was curious about all the things I could do with existential types. Fortunately, my enthusiasm was curbed after reading this blog post on the dangers of type level magic. The author outlines some good and bad uses of existential quantification. He pointed to the foldl package as an example of how to properly employ the technique. It defines the Fold data type:

data Fold a b = forall x. Fold (x -> a -> x) x (x -> b)

If you're unfamiliar with folds, for now just think of them as a more generic form of list reduction.

I decided to implement Fold as well as a tiny portion of the library in TypeScript to really solidify all that I'd learned. Let's walk through it together.

#Implementing Fold in TypeScript

Fold_ is defined just like MkFoo_:

type Fold_<X, A, B> = { step: (x: X, a: A) => X; initial: X; extract: (x: X) => B }

X is the type of the accumulator value. A is the type of the elements in the structure being folded. B is the type of the value resulting from the application of the fold. Many Folds fill in all the variables with the same type. For example, the sum fold specifies them all to be number. But as we'll see, making these type variables distinct enables the creation of more powerful folds.

Next, we move <X> to the right of the =:

type Fold<A, B> = <R>(run: <X>(_: Fold_<X, A, B>) => R) => R

This closely resembles the pattern used when implementing Foo. We're already familiar with the sum fold, so let's implement that now.

const id = <B>(b: B): B => b

const sum: Fold<number, number> = (run) =>
  run({
    step: (x, a) => x + a,
    initial: 0,
    extract: id,
  })

This looks familiar. It's almost exactly the same as the mkFold constructor defined earlier. This makes sense because it constructs a Fold. As mentioned previously, all the types involved in this fold are number, hence Fold<number, number>. By setting initial to 0, the compiler can infer that X is also a number. extract simply returns the accumulator value.

We're still missing one thing: the function that applies a Fold to a structure:

const fold = <A, B>(f: Fold<A, B>, fa: Foldable<A>): B =>
  f(({ step, initial, extract }) => extract(fa.reduce(step, initial)))

Notice how similar this is to the f function defined in the Foo example. It takes a Fold f and applies it to a Foldable fa. Wait, what's a Foldable? You may be wondering why I keep using the word "structure" instead of just saying list. Well, what if we want to fold a linked list or a binary tree? Basically any structure that contains elements of a specific type can be folded. The structure just has to have a reduce method. Conveniently, the [] type already does. Let's define a Foldable interface and then create a new Tree structure that implements it.

type Reduce<A> = <X>(step: (x: X, a: A) => X, init: X) => X

interface Foldable<A> {
  reduce: Reduce<A>
}

In the definition of Reduce, notice how A is on the left of the = and X is on the right. If A and X were on the left then we'd be saying that a Foldable must specify both A and X. If they were on the right, we'd be saying that both A and X can be inferred from step and init. What we want to say is that the structure implementing Foldable should fill in A with the type of its elements and that X is inferred from the arguments. Thus, A on the left, X on the right.

Next, let's define a Tree. A Haskell implementation might look something like this:4

data Tree a = Empty | Leaf a | Node (Tree a) a (Tree a)

This data type is recursive in that a Node, one of the Tree's data constructor, takes a left Tree and a right Tree. TypeScript does not support recursive type aliases but it does have "recursive back references within interface types."5

Unfortunately we can't quite replicate the elegance of the Haskell definition in TypeScript because we need a class on which to define reduce.

type Root<A> = Empty | Leaf<A> | Node<A>

class Tree<A> implements Foldable<A> {
  root: Root<A>

  constructor(root: Root<A> = { tag: 'EMPTY' }) {
    this.root = root
  }

  // reduce: ...
}

interface Empty {
  tag: 'EMPTY'
}

interface Leaf<A> {
  tag: 'LEAF'
  value: A
}

interface Node<A> {
  tag: 'NODE'
  left: Tree<A>
  value: A
  right: Tree<A>
}

A tagged union type is used to describe the Root of the Tree.

  // ...
  reduce: Reduce<A> = (step, init) => {
    const root = this.root
    switch (root.tag) {
      case 'NODE':
        return root.right.reduce(step, step(root.left.reduce(step, init), root.value))
      case 'LEAF':
        return step(init, root.value)
      default:
        // EMPTY
        return init
    }
  }
  // ...

reduce performs a specific action for each of the different root types. The NODE case is clearly the interesting one. It essentially performs an in-order traversal.

Now that we have two different foldable structures, Tree<A> and <A>[], let's test out the sum fold on them.

const testArray = [1, 2, 2, 3]

const testTree = new Tree<number>({
  tag: 'NODE',
  left: new Tree({
    tag: 'NODE',
    left: new Tree({
      tag: 'LEAF',
      value: 1,
    }),
    value: 2,
    right: new Tree(),
  }),
  value: 2,
  right: new Tree({
    tag: 'LEAF',
    value: 3,
  }),
})

fold(sum, testArray) // 8
fold(sum, testTree) // 8

So we've proven that the implementation of Fold can work on various structures but we haven't tried folds that have different concrete types for X, A, and B.

const all = <A>(p: (a: A) => boolean): Fold<A, boolean> => (run) =>
  run({
    step: (x, a) => x && p(a),
    initial: true,
    extract: id,
  })

all takes a predicate p and returns a Fold<A, boolean>. If all the applications of the predicate to each element in the structure result in true, the overall result is true otherwise false. The compiler can infer X from the value of initial and A from the predicate.

const nub = <A>(): Fold<A, A[]> => (run) =>
  run({
    step: (x, a) => x.add(a),
    initial: new Set<A>(),
    extract: (x) => [...x],
  })

nub folds a structure into a list without duplicate elements. X is inferred to be Set<A> from the initial value‌. Because sets prevent duplicates from being added and the TypeScript/JavaScript implementation of Set maintains insertion order6, extract can simply convert the accumulator to a list. Note that this fold is a function even though it doesn't take any arguments. This is because it has a type variable and values cannot have type variables.

Let's put these folds to use:

fold(all((a) => a > 0), testArray) // true
fold(all((a) => a > 0), testTree) // true
fold(all((a) => a < 0), testArray) // false
fold(all((a) => a < 0), testTree) // false

fold(nub(), testArray) // [1, 2, 3]
fold(nub(), testTree) // [1, 2, 3]

All this code can be viewed in the fold-ts repo. Maybe you'll find some use for it. If you'd like to contribute additional folds please submit a PR!

If this type level trickery still confuses you, there are plenty of other examples.7

#Existential Crisis

It's unfortunate that a seemingly trivial problem leads to such a complex solution. So far, the only two "mainstream" languages that have explicit implementations of existential quantification are Haskell and Scala. It would be greatly beneficial for a language like TypeScript, one that is accessible to many, to introduce the concept to the programming masses. It's a powerful feature and one that should be in every developers playbook or at least on their radar.


  1. Existential Type?

  2. Closure Workaround

  3. What's the theoretical basis for existential types?

  4. Foldable

  5. Recursive Back References Within Interface Types

    This example was provided by the creator of TypeScript himself.

  6. Set

  7. Examples