Is there a way to enhance the readability of intellisense output for Typescript generics using the Omit method?

Scenario: I have customized a third-party library for users by removing two properties from all functions in the library. I have implemented a function to achieve this, along with executing a bootstrap function if provided. Here is the code snippet:

const wrapped = <
  F extends (args: any) => any,
  A extends Omit<Parameters<F>[0], "fs" | "http">
>(
  f: F,
  pre?: (params: A) => void
): ((args: A) => ReturnType<F>) => {
  return (args: A): ReturnType<F> => {
    if (pre) {
      pre(args);
    }

    return f({ fs, http, ...args });
  };
};

Everything seems to be functioning correctly, highlighting the appropriate properties.

However, when utilizing a function that has been wrapped in this manner within VSCode, the intellisense output appears cluttered and complex.

To further illustrate this issue with a comprehensive example:

const wrapped = <
  F extends (args: any) => any,
  A extends Omit<Parameters<F>[0], "foo">
>(
  f: F
): ((args: A) => ReturnType<F>) => {
  return (args: A): ReturnType<F> => {
    const foo = "bar"
    return f({ foo, ...args });
  };
};

const thirdPartyFunc = (args: {foo: string; bar: string}) => {
  console.log(args.foo, args.bar);
}

const wrappedThirdPatyFunc = wrapped(thirdPartyFunc);

thirdPartyFunc({foo: "0", bar: "1"})
/**
Intellisense displays simply:

const thirdPartyFunc: (args: {
    foo: string;
    bar: string;
}) => void
*/

wrappedThirdPatyFunc({bar: "1"})
/**
Intellisense appears cluttered (displays Omit, includes omitted keys):

const wrappedThirdPatyFunc: (args: Omit<{
    foo: string;
    bar: string;
}, "foo">) => ReturnType<(args: {
    foo: string;
    bar: string;
}) => void>
 */

I am concerned about the potential negative user experience. Is there a modification I can make to my code to improve the intellisense output and make it resemble the original intellisense, minus the removed fields?

I am unsure of what steps to take next. This project involving generics in TypeScript is the most intricate task I have tackled so far, so I am navigating it cautiously through trial and error.

Answer №1

TypeScript employs a range of heuristic rules to determine how a type should be displayed in IntelliSense quick info. When dealing with a type alias like type Bar<E> = ⋯E⋯, and you use Bar<number>, the compiler must decide whether to show it as-is or substitute number for E in the definition such as ⋯number⋯. If the substituted type itself includes type aliases, further decisions need to be made. While these heuristics generally work well across different scenarios, they may not always align perfectly with the developer's intentions. There have been requests for more flexible or explicit solutions, like microsoft/TypeScript#45954, but currently, developers have to make do with the existing options.

In your situation, if you are encountering the Omit type alias when you don't want to, one approach is to remove the alias entirely and replace it with an inline equivalent. Instead of using its definition which relies on the Pick utility type, followed by inlining that, consider an alternative definition involving key remapping mentioned in microsoft/TypeScript#41383:

type Omit<T, K> = {[P in keyof T as Exclude<P, K>]: T[P]}

This will transform the code into:

const updated = <
    X extends (args: any) => any,
    Y extends {
        [Z in keyof Parameters<X>[0]
        as Exclude<Z, "bar">]: Parameters<X>[0][Z]
    }
>(
    x: X
): ((args: Y) => ReturnType<X>) => {
    return (args: Y): ReturnType<X> => {
        const bar = "baz"
        return x({ bar, ...args });
    };
};

giving you the desired behavior:

updatedThirdPartyFunc({ baz: "1" })
/**
const updatedThirdPartyFunc: (args: {
    baz: string;
}) => ReturnType<(args: {
    bar: string;
    baz: string;
}) => void> */

The above addresses the initial question. To streamline further, I recommend replacing X with the argument types and return types directly:

const updated = <Y extends { bar?: string }, Z>(
    x: (y: Y) => Z
): ((args: { [Q in keyof Y as Exclude<Q, "bar">]: Y[Q] }) => Z) => {
    return args => {
        const bar = "baz"
        return x({ bar, ...args } as Y);
    };
};

const updatedThirdPartyFunc = updated(thirdPartyFunc);

updatedThirdPartyFunc({ baz: "1" })
/**
const const wrappedThirdPatyFunc: (args: {
    baz: string;
}) => void
 */

By eliminating the need for ReturnType and Parameters, the resulting computed and displayed types become simpler.

Link to Playground with Code Example

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