Modify each member of a function overload

Question

Given the following overloaded function:

function fooBar(param1: string, param2: number): number;
function fooBar(param1: boolean): number;
function fooBar(boolOrString: boolean | string, param2?: number): number {
  if (typeof boolOrString === 'string') {
    return boolOrString.length + param2!;
  }
  return boolOrString ? 100 : -100;
}

I would like to provide a type which changes the return type of the function to void without loosing information about the overload.

I've tried something like this:

type VoidFn<F> = F extends ((...args: infer Args) => any) | never
   ? (...args: Args) => void
   : F extends infer F1 | infer F2
   ? VoidFn<F1> | VoidFn<F2>
   : never;

However, it only resolves to the type of the last overload with the void return type:

type FV = VoidFn<typeof fooBar>
// (param1: boolean): void;

Is it possible to create a type which changes all the overloads? I've assumed that overloads are represented as union types in Typescript. Is this assumption correct?

---

Solution

Include the most general overload as the last one, and this is what typescript would pick:

function fooBar(param1: string, param2: number): number;
function fooBar(param1: boolean): number;
function fooBar(boolOrString: boolean | string, param2?: number): number;
function fooBar(boolOrString: boolean | string, param2?: number): number {
  if (typeof boolOrString === 'string') {
    return boolOrString.length + param2!;
  }
  return boolOrString ? 100 : -100;
}

The VoidFn type should be declared simply as:

type VoidFn<F> = F extends ((...args: infer Args) => any) ? ((...args: Args) => void) : never