How to 'map' a Tuple to another Tuple type in Typescript 3.0

Question

I have tuple of Maybe types:

class Maybe<T>{ }

type MaybeTuple = [Maybe<string>, Maybe<number>, Maybe<boolean>];

and I want to turn this into a tuple of types:

type TupleIWant = [string, number, boolean];

so I tried this:

type ExtractTypes<T> = T extends Maybe<infer MaybeTypes>[] ? MaybeTypes : never;

type TypesArray = ExtractTypes<MaybeTuple>; // string | number | boolean NOT [string, number, boolean]

Which doesn't work :-(

I get (string | number | boolean)[] rather than the tuple I want: [string, number, boolean]

Is what I want to do currently possible?

Answer 1

You'll need to use a mapped tuple type, which is supported in TypeScript 3.1.

You can make a mapped type that has properties 0, 1, 2 and length of the correct types, like this:

class Maybe<T> {}

type MaybeTuple = [Maybe<string>, Maybe<number>, Maybe<boolean>];

type MaybeType<T> = T extends Maybe<infer MaybeType> ? MaybeType : never;
type MaybeTypes<Tuple extends [...any[]]> = {
  [Index in keyof Tuple]: MaybeType<Tuple[Index]>;
} & {length: Tuple['length']};

let extractedTypes: MaybeTypes<MaybeTuple> = ['hello', 3, true];

If you're using an older version of typescript, you can use an unreleased version of TypeScript, or as a workaround, you can write a conditional type to match against tuples as long as you think you are likely to have, e.g., like this.

Answer 2

You need to exclude the non-index keys of the array, or you'll also be mapping over push, length, etc:

type NumStrDate = [number, string, Date]
type IdxOf<T extends any[]> = Exclude<keyof T, keyof any[]>
type OrUndef<T extends any[]> = {[K in IdxOf<T>]: T[K] | undefined}
type MaybeNSD = OrUndef<NumStrDate>

const fine: MaybeNSD = [5, 'foo', undefined]
const problem: MaybeNSD = ['foo', undefined, undefined]

If you hover on MaybeNSD it will show you the resolved meaning:

type MaybeNSD = {
    0: number | undefined;
    1: string | undefined;
    2: Date | undefined;
}

playground link