How can we prevent transform Omit into Pick in Typescript when utilizing generics?

Is there a way to prevent this behavior? It's crucial when compiling type declarations solely for libraries.

I specifically require it for compiling declarations only:

tsc --declaration true --emitDeclarationOnly true

Here is a minimal example of the code:

function test<T>(t: T): Omit<T, 'href'> {
  return t;
}

export const TEST_CONST = test({} as HTMLAnchorElement);

Expected output:

export declare const TEST_CONST: Omit<HTMLAnchorElement, 'href'>

Received output:

export declare const TEST_CONST: Pick<HTMLAnchorElement, "toString" | "search" | ... (truncated for brevity) ... | "protocol" | "username">;

An explicit type declaration will function, but can be cumbersome with lengthy chains of function calls:

function test<T>(t: T): Omit<T, 'href'> {
  return t;
}

export const TEST_CONST: Omit<T, 'href'> = test({} as HTMLAnchorElement);

Answer №1

It is to be anticipated and considered a valid value. This occurs because Omit is essentially a form of Pick combined with Exclude.

/**
 * Create a type containing all properties of T except those specified in K.
 */
type Omit<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;

You can try defining your own type for this scenario.

type NoHref<T> = Omit<T, 'href'>;

function test<T>(t: T): NoHref<T> {
  return t;
}

export const TEST_CONST = test({} as HTMLAnchorElement);

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