fix(lib/es5): Preserve Omit in output

[ExE-Boss]

I’ve been using this without problems in my local version of the TypeScript library.

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Fixes #34793

[dragomirtitian]

I am pretty sure this option was considered when Omit was added. It was rejected because it does not preserve optionality and readonly-ness. Pick is homomorphic since it maps over K where K extends keyof T and thus preserves modifiers.

Since your type maps over Exclude<keyof T, K> it is not homomorphic. Only types that map over keyof T where T is a type parameter or that map over K where K is type parameter with K extends keyof T are homomorphic.

type Omit2<T, K extends keyof any> = {
    [P in Exclude<keyof T, K>]: T[P];
};

type U = {
    a: number,
    readonly b?: number
} 

type R = Omit2<U, 'a'> // has b: number | undefined;
type R2 = Omit<U, 'a'> // has  readonly b?: number | undefined;

Playground Link

[ExE-Boss]

That is probably also why auto‑fix is broken with this.

[ypresto]

FYI:

The documentation for "homomorphic" and Pick is here: https://www.typescriptlang.org/docs/handbook/advanced-types.html#mapped-types

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#35889 (comment)
type R = Omit2<U, 'a'> // has b: number | undefined; is actually // has b: number; (optional is NOT copied too).

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I tried

type Omit2<T, K extends keyof any> = K extends keyof T ? {
  [P in Exclude<keyof T, K>]: T[P];
} : Omit<T, K>;

but does not solve the problem 😂

[ypresto]

Then my question is: why Omit uses K extends keyof any instead of K extends keyof T? (I sometimes specify wrong or unnecessary K for Omit)

[ExE-Boss]

The reason is probably because this is the more correct Object.assign(…) type:

declare function __assign<T, S>(target: T, source: S): Omit<T, keyof S> & S;

To be 100% correct, the result should include properties from T that are optional in S.