Apply array operator distributively in TypeScript

Question

Let's say I have a type which is a union of literals:

type X = "a" | "b" | "c";

Now I want to introduce a new type signifying a union of arrays of each type in X. In other words, I want to apply the array operator distributively over the union.

Just adding the array operator on the type doesn't do what I want – this will permit a mixed array e.g.

type Y = X[]; // ("a" | "b" | "c")[]
const xs: Y = ["a", "b"]; // valid

However, conditional types have the distributive property that I seek. I seem to be able to achieve my goal by using them in conjunction with a condition that never gets fulfilled:

type Arrayify<T> = T extends never ? never : T[];
type Y = Arrayify<X>; // "a"[] | "b"[] | "c"[]
const xs: Y = ["a", "b"]; // invalid

As far as I can tell it solves my problem, but it feels a little backwards to take the detour through conditional types only to get the distributivity. Is there a better, more conventional way?

Titian Cernicova-Dragomir · Accepted Answer · 2019-07-25 08:10:09Z

3

Your approach is the right one. Using conditional types for distribution is the way to accomplish this. The recommended patterns to get just the distributive behavior is T extends T or T extends unknown, but yours works just fine as well in this case.

type Arrayify<T> = T extends T ? T[] : never;

answered Jul 25, 2019 at 8:10

Titian Cernicova-Dragomir

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2 Comments

Hannes Petri Over a year ago

I just realised there's a small complication applying this approach to booleans, since they are defined as true | false unions. Thus, Arrayify<boolean> will result in true[] | false[] and not boolean[]. Is there any way of "blocking" the distributivity?

Titian Cernicova-Dragomir Over a year ago

@HannesPetri the usual way to block distributivity is to use a tuple. This should work : type Arrayify<T> = [T] extends [boolean] ? T[] : T extends T ? T[] : never; although you will still get strange distribution over enums.

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