TypeScript：如何包装函数，更改其返回类型？

Question

I need to write a function like this:

const wrapper = (fn) => () => {
  const value = fn.apply (this, arguments)
  const somethingElseEntirely: WellDefinedType = doMagic (value)
  return somethingElseEntirely
}

...that wraps any given function. It is known that given function returns a well defined type, say, string . It is also known that given function can accept any combination of arguments, and the wrapped function should take the same arguments, however , it should return a different type of value.

Eg, a function like:

function foo (arg1: string, arg2?: number): string

...should become:

(arg1: string, arg2?: number): WellDefinedType

Is this possible to achieve in TypeScript without resorting to any ?

Answer 1

TypeScript 3.0 recently introduced new features allowing you do exactly this.

declare function wrapWithNumberReturn<A extends any[]>(fn: (...args: A) => any): (...args: A) => number

declare const concat: (a: string, b?: string) => string

// returns (a: string, b?: string) => number
const wrapped = wrapWithNumberReturn(concat)

Playground here