TypeScript mapping over unboxed conditional: turn OR into AND

Question

I have an array that contains multiple objects with functions foo.

Now I want to construct a new object signature with a function foo that inherits all signatures from the array item foo functions.

let arr = [
    { foo: (a: 'a') => 'A' as const },
    { foo: (a: 'b') => 'B' as const },
];

type MapAndUnion<T extends ReadonlyArray<any>> = { foo: T[number] extends { foo: infer V } ? V : never }

type U = MapAndUnion<typeof arr>

unfortunately, I am getting

type U = {
    foo: ((a: "a") => "A") | ((a: "b") => "B");
}

This is not callable, as the signatures are conflicting.

Is there a way to get (AND instead of OR)

type U = {
    foo: ((a: "a") => "A") & ((a: "b") => "B");
}

?

playground

Possible duplicate of Transform union type to intersection type — jcalz
– jcalz, Commented Jan 3, 2020 at 20:01

jcalz · Accepted Answer · 2020-01-03 20:05:41Z

3

You can use type inference in conditional types in a contravariant position to get intersections instead of unions:

type MapAndIntersection<T extends ReadonlyArray<any>> =
  { foo: ((x: T[number]) => void) extends ((x: { foo: infer V }) => void) ? V : never }

type U = MapAndIntersection<typeof arr>
// type U = { foo: ((a: "a") => "A") & ((a: "b") => "B"); }

Hope that helps; good luck!

Link to code

answered Jan 3, 2020 at 20:05

jcalz

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