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,
type Fn = (a: number, b: string) => number
type Result = AppendArgument<Fn, boolean>
// expected be (a: number, b: string, x: boolean) => number
This question is ported from the original article by @maciejsikora
import type { 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>>,
// @ts-expect-error
AppendArgument<unknown, undefined>,
]