Ability to replace return type of function type

Search Terms

• replace return type
• parameters generic

Suggestion

I would like to be able to replace just the return type of a function with a utility type, like WithReturn<F, R>.

I'm not 100% sure this is needed, but it seems like using (...args: Parameters<F>) => R is not sufficient because any generics in F are lost.

Use Cases

My use case is a utility function that captures the arguments to a function to allow it to be called later.

Examples

export const defer = <F extends (...args: any) => any>(f: F) => ((
  ...args: Parameters<F>
) => ({
  function: f,
  args,
})) /* as any as F */;

const map = <T, R>(array: Array<T>, f: (i: T) => R) => array.map(f);
const deferredMap = defer(map);

let r = deferredMap([1], (i: number) => 2); // error now: Type 'unknown' is not assignable to type 'number'.(2345)
r.args; // error if defer returns F

Right now there's an error because the generics to map() are lost when using Parameters.

Playground: https://www.typescriptlang.org/play/?ssl=13&ssc=1&pln=1&pc=1#code/KYDwDg9gTgLgBAYwgOwM7wCbAGbCnAXjgB4AxOUGYZDVOACgDpmBDKAc1QC44XkBPAJSEAfLwEj62HqWEEx9egCg4cZozaceABTYsAtsCpRUZEUrkKA3irjYArsgQwAlih7YANLc2pvAX0FhAHoAKl46Pn4IuHJQ4IBuJSUkNHh9FjBCEgAVTzgAJUk2KBZ+HgBBKFL+YhyRfOkGFx4cy0L2krLGDLApQSTU9DgsXGrgDABZTOzRvHpegeSAGyM4fCI58anM+gBtAEYAXXz6Frhke30AIzx2gCYlqA0OVCSgA

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

リンクされた TypeScript Playground でジェネリック消失の挙動を再現することから始め、Parameters と提案されている WithReturn<F, R> の挙動に焦点を当てます。関数型のジェネリックパラメーターを保持したまま戻り値の型だけを置き換え、JavaScript のランタイム出力を変更しない Utility が実現できれば完了です。