Generic type gets widened in an unexpected way

[Harpush]

Bug Report

🔎 Search Terms

widen

🕗 Version & Regression Information

4.9.4

⏯ Playground Link

Playground link with relevant code

💻 Code

enum One {
  A = 'a',
  B = 'b',
  C = 'c'
}

const isOneSomethingMap = {
  [One.A]: true,
  [One.B]: false,
  [One.C]: true
} as const satisfies Record<One, boolean>;

type BooleanMapToUnion<T extends Record<string, boolean>> = {
  [P in keyof T]: T[P] extends true ? P : never;
}[keyof T];

type SomethingOne = BooleanMapToUnion<typeof isOneSomethingMap>;

const a = <T>(value: T) => value;
const b = <T>(fn: (value: T) => T, v: T): T => fn(v);

const v: SomethingOne = One.A as SomethingOne;
const v2: One.A | One.C = One.A as One.A | One.C;

const r1 = b(a, v); // One and not SomethingOne - not expected
const r2 = a(v); // SomethingOne as expected
const r3 = b(a, v2); // Union as expected

const r4 = b(a, One.A as One.A | One.C); // Works
const r5 = b(a, One.A as SomethingOne); // Still doesn't work...

🙁 Actual behavior

The type gets widened when using BooleanMapToUnion... I guess it is the culprit.

🙂 Expected behavior

Both union and using BooleanMapToUnion should work equally.

[whzx5byb]

In ts 5.0 you can just write const b = <const T>(fn: (value: T) => T, v: T): T => fn(v); (Playground).

See #51865 for details.

[Harpush]

@whzx5byb It's indeed better but:

1. You lose the original type SomethingOne and get union instead.
2. The original problem came from rxjs types which I have zero control over.

Adding another weird case and easier one (again works for union and not BooleanMapToUnion):

const e = <T>(value: T): {a: T} => ({a: value});
const g = e(v); // Gets widened
const e2 = <T>(value: T): T => value;
const g2 = e2(v); // Works

[whzx5byb]

The widening behavior is intended, see #10676.

And if you want to preserve the type alias you can just add & {} just like type SomethingOne = BooleanMapToUnion<typeof isOneSomethingMap> & {}, according to #31940 (comment).

[Harpush]

With all respect to the new const T feature I think using it here just masks the bug.
There is no reason using type SomethingOne = One.A | One.C will behave correctly while BoooleanMapToUnion won't.
const T will force specific types and probably libraries like rxjs won't change everything to it thus the problem remains.

Considering & {} - Seems like that solves it even without const T... Thats certainly weird though.

[fatcerberus]

I’m pretty sure that BooleanMapToUnion<typeof isOneSomethingMap> doesn’t actually produce One.A | One.C. Setting a property to never in the mapped type doesn’t remove it. You need to use an as clause to rename the key to never to remove it.

[Harpush]

I’m pretty sure that BooleanMapToUnion<typeof isOneSomethingMap> doesn’t actually produce One.A | One.C. Setting a property to never in the mapped type doesn’t remove it. You need to use an as clause to rename the key to never to remove it.

That's exactly what I am doing though...

[RyanCavanaugh]

I'd agree something weird appears to be happening here. A demonstration is to slightly tweak these definitions

const b = <T>(fn: null | ((value: T) => T), v: T): T => fn!(v);
const r1 = b(null, v); // SomethingOne, as expected

The call of b(a, v) where a is the identity function shouldn't have changed the inference to One here; I don't see a good reason for that to have happened.