Generic, optional type as an optional argument in TypeScript

April 14, 2023

How can I create an optional function argument who’s type is generic?

I have the following type definition:

type Ok<T> = { ok: true; value: T }

I want to write a function that combines the behavior of the two following functions into one:

const Ok = <T>(value: T): Ok<T> => {
    return { ok: true, value }
}
const Ok = (): Ok<undefined> => {
    return { ok: true, value: undefined }
}

The argument is optional.
If the argument is not included, then the generic type is undefined.
If the argument is included, then the generic type is the type of the argument.

So you’d use the function like the following:

const emptyOk = Ok() // type Ok<undefined>
const okOne = Ok(1) // type Ok<number>

The end goal is to emulate Rust’s Result type. I know there’s several existing npm packages for emulating Rust’s Result type. I’m looking to do something slightly different from any I found. I’ve also simplified things here to get to the root of my issue with TypeScript.

>Solution :

You can create an overloaded function like this:

function Ok<T extends undefined>(value?: T): Ok<T>
function Ok<T>(value: T): Ok<T>;
function Ok<T>(value: T): Ok<T> {
    return { ok: true, value }
}

The overload for the undefined type prevents type issues caused by allowing the property to always be undefined.

Demo: https://tsplay.dev/mqY0Jm