Typescript inference from arguments

Question

I'm looking for a way to make Typescript guess the return type of my function according to the function arguments.

function fn<T extends object>(a: T, b: T, property: keyof T): any {
  return a[property] ?? b[property];
}

I want to remove any to get the proper return type.

interface A {
  foo?: string;
}

const a: A = { foo : 'bar' };
const b: A = {};
const a = fn(a, b, 'foo'); // it should get the string type from inference

I looked at ReturnType<T> by using it like this ReturnType<typeof T[property]> but it seems not supported by Typescript. I don't known if it's feasible?

kaya3 · Accepted Answer · 2019-12-23 15:36:07Z

2

Add another type parameter for the property name:

function fn<T, K extends keyof T>(a: T, b: T, property: K): T[K] {
  return a[property] ?? b[property];
}

Note that it will infer string | undefined, not string, because the foo property on the interface A is optional, so it's possible both a and b don't have it.

Playground Link

answered Dec 23, 2019 at 15:36

kaya3

51.6k7 gold badges87 silver badges118 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Typescript inference from arguments

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related