TS function return type inference with an optional parameter

Question

I think an example is the better way to explain:

const fn = (fallback?: string) => someFnReturningStringOrUndefined() || fallback;

const result = fn('fallback'); //inferred type for `result` should be `string`, why it is `string | undefined`

If I didn't pass the fallback parameter I would understand.

Answer 1

There are limits to TypeScript's code path analysis, although I think the inference of the return type will always be a union of the possible result types, even if the code path is really explicit.

If you want the result you've described, the easiest way to do it is function overloads:

function fn(a: string): string | undefined;
function fn(a: string, b: string): string;
function fn(a: string, b?: string): string | undefined {
    return a || b;
}
const result1 = fn("", "fallback");
//    ^? const result1: string
const result2 = fn("");
//    ^? const result2: string | undefined

Playground example

Or with an arrow function:

type Fn = {
    (a: string): string | undefined;
    (a: string, b: string): string;
};
const fn = ((a: string, b?: string) => a || b) as Fn;
const result1 = fn("", "fallback");
//    ^? const result1: string
const result2 = fn("");
//    ^? const result2: string | undefined

Playground example