Infer the return type function based on optional parameters

Question

    // Silly function that does nothing
    function f(a: number, b?: number[], c?: number): string | boolean {
        if (b === undefined) 
            return false;

        b.push(a);
        if (c) b.push(c);
        return b.toString();
    }
    const boolTypeValue = f(5);                // type: boolean | string
    const boolTypeValue = f(5, undefined, 8);  // type: boolean | string
    const stringValue = f(9, [], 0);           // type: boolean | string

Is it possible to define f() to infer the return type based on the second optional parameter value, maintaining the order of the parameters.

Answer 1

Use overloads to easily have the return type depend on the parameter type:

function f(a: number, b?: undefined, c?: number): false;
function f(a: number, b: number[], c?: number): string;

function f(a: number, b?: number[], c?: number): string | false {
    if (b === undefined) 
        return false;

    b.push(a);
    if (c) b.push(c);
    return b.toString();
}

const boolTypeValue: boolean = f(5);
const boolTypeValue2: boolean = f(5, undefined, 8);
const stringTypeValue: string = f(9, [], 0);

If you also want to be able to pass a number[] | undefined number[] | undefined value, you need a third overload for that:

function f(a: number, b?: number[], c?: number): string | false;

declare const possiblyUndefinedArray: number[] | undefined;
const boolOrStringTypeValue: string | false = f(9, possiblyUndefinedArray, 0);