Make use of the Object Property method and leverage the strong type inference capabilities of TypeScript to enhance the functionality

My goal is to write code that utilizes Object methods properly and ensures accurate function inference.

  const P = <T,>(x: T) => ({ "foo": (R: Function) => R(x) });

  const f = (a: number) => a + 1;
  const g = (a: number) => a.toString();

  const p1 = f(5); //const p1: number         // good
  const p2 = g(5); //const p2: string         // good

  const q1 = P(5)['foo'](f); //const q1: any   // bad
  const q2 = P(5)['foo'](g); //const q2: any   // bad

  console.log(q1);
  console.log(q2);

TypeScript playground link

Any suggestions or insights are appreciated. Thank you!

"use strict";
const P = (x) => ({ "foo": (R) => R(x) });

const f = (a) => a + 1;
const g = (a) => a.toString();

const p1 = f(5); //const p1: number         // good
const p2 = g(5); //const p2: string         // good

const q1 = P(5)['foo'](f); //const q1: any   // bad
const q2 = P(5)['foo'](g); //const q2: any   // bad

console.log(q1);
console.log(q2);

Answer №1

Utilizing the Function type might not be the best approach; it represents an untyped function that can accept any parameters and returns the unsafe any type. This leads to P being inferred with this type:

const P = <T,>(x: T) => ({ "foo": (R: Function) => R(x) });
/* const P: <T>(x: T) => {
  foo: (R: Function) => any;
} */

Therefore, P(x).foo(y) will consistently return a value of type any for all x and y. Whoops.


If you want the compiler to manage both the type of x and the return type of R, you need to make the returned function generic and provide an appropriate function type for R, like so:

const P = <T,>(x: T) => ({ "foo": <U,>(R: (x: T) => U) => R(x) });

Now it should work as intended:

const q1 = P(5)['foo'](f);
//const q1: number
console.log(q1.toFixed(2)) // "6.00";
const q2 = P(5)['foo'](g);
//const q2: string
console.log(q2.repeat(2)) // "55";

Playground link to code

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