Split tuple of unions into union of tuples in TypeScript

Question

This question is similar to TypeScript extract union type from tuple , but instead of asking why one isn't assignable to the other (I understand that bit), I want to know how to make a function or type split ['a' | 'b', number] ['a' | 'b', number] into ['a', number] | ['b', number] ['a', number] | ['b', number] .

Suppose I have this contrived code:

// the return type of this needs to change, and ideally it can infer F and S without 
// specifying them the function call
function treatAsConstants<
  F,
  S,
  T extends [F, S][] = [F, S][]
>(array: T): (readonly [F, T[number][1]])[] {
  return array;
}

const array = treatAsConstants<'a' | 'b', number>([['a', 1], ['b', 2]]);

// desired: (readonly ['a', number] | readonly ['b', number])[]
// actual: (readonly ['a' | 'b', number])[]
type A = typeof array;

In a later function that takes in array , I need to Extract ['a', number] and ['b', number] in order to determine the shape of an object. However, ['a' | 'b', number] ['a' | 'b', number] fits neither of those types.

Here's my best attempt at splitting the unions, but it just gives me exactly what I started with.

type SplitUnions<
  TKey extends string,
  TVal,
  O extends { [K in TKey]: [K, TVal] } = { [K in TKey]: [K, TVal] }
> = [O[TKey][0], O[TKey][1]];

type Split = SplitUnions<'a' | 'b', number>;

Answer 1

I was pretty close. Instead of [O[TKey][0], O[TKey][1]] I simply needed O[TKey] .