Create union type based on array of types

Question

Is it possible to define union type like this:

Union<T extends any[]> = // something...

and would be used like

Union<[string, number]> // would create (string | number)

I know I can immediatelly define union above, but I need it for usage in generics where it would end up being something like Union<[TDynamic, UDynamic]>

Ingo Bürk · Accepted Answer · 2019-10-06 08:55:29Z

1

You can use the following type:

type Union<T extends any[]> = T extends (infer U)[] ? U : never;

This is actually basically the same type as Unpacked defined in the docs, and works because a type tuple becomes a type union in this case.

type X1 = Union<[number, string]>; // string | number
type X2 = Union<[number, number]>; // number
type X3 = Union<[number]>; // number

If you're on an older TypeScript version which does not support infer, you can also use:

type Union<T extends any[]> = T[number];

answered Oct 6, 2019 at 8:55

Ingo Bürk

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4 Comments

zhuber Over a year ago

Perfect, any way to be able to define like Union<number, string> (to support dynamic number of generic arguments so I don't have to add array definition)?

Ingo Bürk Over a year ago

If you can write Union<T1, T2>, why not write T1 | T2 directly? I don't think for that you need a separate type.

zhuber Over a year ago

Because I'll use it in generic definition but I get your point, because I won't be able to pass it like that anyway in generic definition (will have array of dynamic types). One more question if you don't mind - is it possible to create Intersection type (same way as above Union)?

Ingo Bürk Over a year ago

You can possibly get that to work using @jcalz's UnionToIntersection trick: stackoverflow.com/a/50375286/1675492

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