get `keyof` non-optional property names in typescript

Question

This is my interface

interface X {
    key: string
    value: number | undefined
    default?: number
}

But I want the non-optional keys only, aka. "key" | "value", or just "key" (both will do fine for me)

type KeyOfX = keyof X gives me "key" | "value" | "default".

type NonOptionalX = {
    [P in keyof X]-?: X[P]
}

type NonOptionalKeyOfX = keyof NonOptionalX gives "key" | "value" | "default" as -? only removes the optional modifier and make all of them non-optional.

ps. I use Typescript 2.9.

Answer 1

You can use conditional type operator of the form undefined extends k ? never : k to substitute never for keys of values that undefined could be assigned to, and then use the fact that union T | never is just T for any type:

interface X {
    key: string
    value: number | undefined
    default?: number
}


type NonOptionalKeys<T> = { [k in keyof T]-?: undefined extends T[k] ? never : k }[keyof T];

type Z = NonOptionalKeys<X>; // just 'key'

Also this comment might be relevant: https://github.com/Microsoft/TypeScript/issues/12215#issuecomment-307871458