Creating an exhaustive array from a union type

[gxxcastillo]

Say I have:

type T = 'one' | 'two' | 'three'

It would be great to generate a type, E that requires all union values, which in this case would be something like this but without enforcing order:

type E = ['one', 'two', 'three']

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[fregante]

What for?

without enforcing order

I don’t think that's possible, the type you suggested is a tuple, which is ordered. I suppose extends can be used to build a type like if x[0] is 'one', then x[1] must be 'two' | 'three', but that sounds extremely verbose and complex.

[sindresorhus]

I think you could use the #686 type for this when it lands.

Untested ChatGPT generated:

type T = 'one' | 'two' | 'three';

type AllMembersPresent<Union> = UnionToTuple<Union> extends infer Array
	? (Array extends UnionToTuple<Union>
		? (Union extends Array[number] ? Array : never)
		: never)
	: never;

type E = AllMembersPresent<T>;

const validArray: E = ['one', 'three', 'two']; // This is valid
const invalidArray1: E = ['one', 'three'];     // This will throw a type error
const invalidArray2: E = ['one', 'three', 'four']; // This will throw a type error