O.Optional key param has changed, using any breaks the type

[tonivj5]

🐞 Bug Report

Describe the bug

In 6.13.36 this type worked right using any as Key param, now it doesn't

Reproduce the bug

type Test = O.Optional<{ test: { subtest: string } }, any, 'deep'>;
const test: Test = null;
// test has this type
// const test: {
//   [x: string]: OptionalDeep<any>;
// }

Expected behavior

Expected type (as in 6.13.36):

const test: {
    test?: OptionalDeep<{
        subtest: string;
    }>;
}

Possible Solution

Using A.Key as Key param works 👍

type Test = O.Optional<{ test: { subtest: string } }, A.Key, 'deep'>;

Screenshots

Additional context

I'm using latest typescript version (3.9.7).

This worked before, but maybe this is the expected behavior 😅

[millsp]

Hey @tonivj5, thanks for reporting this, I know. The reason that this was working before is a bug that I corrected. This now behaves like any other mapped type should. Sorry about this! And yes, you should always use A.Key:

type t0 = Pick<{a: 1, b: 2}, any> // {[x: string]: any}

Cheers!