Refer: Medium - Append Argument

Describe

For given function type Fn, and any type A (any in this context means we don’t restrict the type, and I don’t have in mind any type 😉) create a generic type which will take Fn as the first argument, A as the second, and will produce function type G which will be the same as Fn but with appended argument A as a last one.

For example,

code snippetCopytypescript
type Fn = (a: number, b: string) => number

type Result = AppendArgument<Fn, boolean> // expected be (a: number, b: string, x: boolean) => number

Test Cases

code snippetCopytypescript
import { Equal, Expect } from '@type-challenges/utils'

type Case1 = AppendArgument<(a: number, b: string) => number, boolean>
type Result1 = (a: number, b: string, x: boolean) => number

type Case2 = AppendArgument<() => void, undefined>
type Result2 = (x: undefined) => void

type cases = [Expect<Equal<Case1, Result1>>, Expect<Equal<Case2, Result2>>]

Solution

code snippetCopytypescript
type AppendArgument<Fn extends (...arg: any[]) => any, A> = Fn extends (...arg: infer P) => any
  ? (...arg: [...P, A]) => ReturnType<Fn>
  : never