Typescript copying only function arguments and not return type

Question

I have some util methods and I'm wondering if there's a way to copy arguments from one method to another. I was playing around with typeof and trying to type the 2nd function that way but I can't quite figure it out.

declare function foo(a: number, b: string): number;

now I want a type bar to have foo's arguments, but not the return type, for example let's say it calls foo but doesn't return anything:

const bar = (...args) => { foo(...args); }

Now I can declare bar to have the exact same type as foo:

const bar: typeof foo = (...args) => { foo(...args); }

but the return type doesn't match now. So how do I either:

• Just copy the argument signature
• Change the return type I get from typeof foo

Answer 1

There's built-in Parameters type

declare function foo(a: number, b: string): number;

type fooParameters = Parameters<typeof foo>;

declare const bar: (...parameters: fooParameters) => void;
// inferred as const bar: (a: number, b: string) => void