Typescript: is there a type map for arrays

Issue

Say I have a list of booleans:

const a: boolean[] = [true, false, false];

Can I build a type conditional that fires only when at least one of the booleans is true?

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I can do it for the case where they're all true:

type BList = true[] | false[];

type AreAllTrue<T extends BList> = T[number] extends true ? true : false;

const a: BList = [true, true, true];

const b: AreAllTrue<typeof a> = true; // This compiles
const c: AreAllTrue<typeof a> = false; // This doesn't (as expected)

// type AnyAreTrue<???> = ???

But I need the list to have both true and false, not just a list full of one type of boolean.

Solution

Perhaps this is what you want?

type AnyIsTrue<T extends boolean[]> = T extends false[] ? false : true

type _test0 = AnyIsTrue<[]>  // false
type _test1 = AnyIsTrue<[false]>  // false
type _test2 = AnyIsTrue<[false, true]>  // true
type _test3 = AnyIsTrue<[false, true, false]>  // true

Hopefully this makes sense: an array of booleans doesn't have any trues if and only if it's a subtype of false[].

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However, as I said in the comment, if the type of your variable is boolean[], TypeScript can't say much about it. It doesn't know what values are in the array.

type _test4 = AnyIsTrue<boolean[]>  // true

Answered By - decorator-factory