Omit returns an invalid type when the target has a generic key

[szmarczak]

Bug Report

🔎 Search Terms

omit generic

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about https://github.com/Microsoft/TypeScript/wiki/FAQ#common-bugs-that-arent-bugs

⏯ Playground Link

Playground link with relevant code

💻 Code

class Example {
    foo: number;
    bar: string;
	baz: string;

    constructor() {
        this.foo = 1;
        this.bar = '';
		this.baz = '';
    }

    [key: string]: unknown;
}

const test: Omit<Example, 'bar' | 'baz'> = new Example();

// `foo` is now `unknown`
console.log(test.foo + 5);

🙁 Actual behavior

test.foo is unknown

🙂 Expected behavior

test.foo should be number

[jcalz]

Duplicate of #43139 and others

[szmarczak]

This is extremely counterintuitive.

[MartinJohns]

It's due to the design of the language, and there are workaround available.

[szmarczak]

I've updated the example. OmitIndex doesn't support union types as keys to remove.

[szmarczak]

there are workaround available.

I'm sure there are, but I don't want to write long lines like this

OmitIndex<OmitIndex<OmitIndex<Example, 'foo'>, 'bar'>, 'baz'>

[MartinJohns]

I've updated the example. OmitIndex doesn't support union types as keys to remove.

There is no mention of a OmitIndex in your example.

[szmarczak]

See #41966

type EqualsTest<T> = <A>() => A extends T ? 1 : 0;
type Equals<A1, A2> = EqualsTest<A2> extends EqualsTest<A1> ? 1 : 0;

type Filter<K, I> = Equals<K, I> extends 1 ? never : K;

type OmitIndex<T, I extends string | number> = {
  [K in keyof T as Filter<K, I>]: T[K];
};

[szmarczak]

Figured it out. I just need to fix the Filter like this:

type Filter<K, I> = Equals<K, I> extends 1 ? never : IfPrimitiveString<I, K, (K extends I ? never : K)>;

// or (which I prefer)
type Filter<K, I> = IfPrimitiveString<I, Equals<K, I> extends 1 ? never : K, K extends I ? never : K>;