Playing with contravariant functors in typescript

Lets define a sum type Ordering with its data constructors:

class LT {}
class EQ {}
class GT {}
type Ordering = LT | EQ | GT;
class Comparison<A> {
  constructor(public runComparison: ((a: A) => (b: A) => Ordering)){}
}

Because of the position of the parameter we can see that this is a contravariant structure

Example usage:

• We can compare numbers

const x = new Comparison<number>((a) => b => {
  if (a === b) return new EQ();
  if (a > b) return new GT();
  if (a < b) return new LT();
})

Now lets define our contravariant functor over Comparison.

As you know functors map has a type of:

<A, B>(f: (a: A) => B) => (a: F A) => F B. 

And while functors pack things (think of array, or function composition) the contravariant functors "unpack".
Its type is:

<A, B>(f: (b: B) => A) => (a: F A) => F B

For Comparison this is:

  type contramapT = <A, B>(f: (b: B) => A) => (a: Comparison<A>) => Comparison<B>;
  const contramap: contramapT = (f) => compC => {
    return new Comparison(a => b => compC.runComparison(f(a))(f(b)))
  }

Let say we have some users which age we would like to compare.

  const compareNumbers = new Comparison<number>((a) => b => {
    if (a === b) return new EQ();
    if (a > b) return new GT();
    if (a < b) return new LT();
  });

  type userT = {
    name: string;
    age: number;
  }

  const user1 = { name: "Igor", age: 25 }
  const user2 = { name: "Ivan", age: 28 }

  const userAgeComparer = contramap<number, userT>(a => {
    return a.age;
  })(compareNumbers)
  const result = userAgeComparer.runComparison(user1)(user2);

As you can see from this trivial example we generated a new comparer for our user.