Implement an overloaded callable interface in typescript

Question

The discussion in implement callable interface is useful but does not completely my answer my question.

interface lol {
    (a: number): (b: number) => string
    // (a: string): (b: string) => string  // overloaded will not work
}
let l: lol = function(a: number) {
  return (b: number) => {
    return (a + b).toString()
  }
}

This snippet works but when the overloaded function is uncommented it fails. I've tried various methods and haven't figure out how to implement the overloaded version.

A slightly convoluted but real world implementation I'm trying to understand is fp-ts style interfaces. Ignoring the complex types it's an overloaded callable interface is what I understand. How would such a thing have a concrete implementation.

export interface Traverse1<T extends URIS> {
  <F extends URIS4>(F: Applicative4<F>): <A, S, R, E, B>(
    ta: Kind<T, A>,
    f: (a: A) => Kind4<F, S, R, E, B>
  ) => Kind4<F, S, R, E, Kind<T, B>>
  <F extends URIS3>(F: Applicative3<F>): <A, R, E, B>(
    ta: Kind<T, A>,
    f: (a: A) => Kind3<F, R, E, B>
  ) => Kind3<F, R, E, Kind<T, B>>
  <F extends URIS3, E>(F: Applicative3C<F, E>): <A, R, B>(
    ta: Kind<T, A>,
    f: (a: A) => Kind3<F, R, E, B>
  ) => Kind3<F, R, E, Kind<T, B>>
  <F extends URIS2>(F: Applicative2<F>): <A, E, B>(
    ta: Kind<T, A>,
    f: (a: A) => Kind2<F, E, B>
  ) => Kind2<F, E, Kind<T, B>>
  <F extends URIS2, E>(F: Applicative2C<F, E>): <A, B>(
    ta: Kind<T, A>,
    f: (a: A) => Kind2<F, E, B>
  ) => Kind2<F, E, Kind<T, B>>
  <F extends URIS>(F: Applicative1<F>): <A, B>(ta: Kind<T, A>, f: (a: A) => Kind<F, B>) => Kind<F, Kind<T, B>>
  <F>(F: Applicative<F>): <A, B>(ta: Kind<T, A>, f: (a: A) => HKT<F, B>) => HKT<F, Kind<T, B>>
}

spender · Answer 1 · 2022-03-23T17:20:33.303

1

The implementation declaration must satisfy all of the overloads stated by the interface so, in your case, it might end up looking something like this:

let l: lol = function <T extends number | string>(a: T): (b: T) => string {
    throw Error("not implemented");
}

In my experience, you'll need to do a lot of type-checking and casting to get the implementation right.

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edited Mar 23 '22 at 17:20

answered Mar 23 '22 at 17:14

spender

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Aah now I get it. Atleast in the `fp-ts` example by using `` some overloads can be eliminated and not need to be implemented. Adding possible solution as separate answer. – twitu Mar 23 '22 at 17:35

score 0 · Answer 2 · answered Mar 23 '22 at 17:38

After @spender's helpful explanation I realized what the fp-ts style interface is actually doing. By having <F extends ...> check it is possible to resolve the overloaded functions to a single concrete implementation as in the example below.

interface lol {
    <T extends number>(a: T): (b: T) => string
    <T extends string>(a: T): (b: T) => string
}

let l: lol = function(a: number) {
    return (b: number) => {
        return (a + b).toString()
    }
}

This instantiates the T as number so the <T extends string> overloading is never resolved (?? is that the right word for it).

Implement an overloaded callable interface in typescript

2 Answers2