Append Argument

In this medium-level challenge, you'll implement the AppendArgument<Fn, A> type, that appends a new argument to parameters of a function type. This can be done with built-ins but it's a good exercise to understand how to manipulate function types in TypeScript.

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

Detailed Explanation

type AppendArgument<Fn extends (...args: any[]) => any, A> = (
  ...args: [...Parameters<Fn>, x: A]
) => ReturnType<Fn>

How it works:

• First we constrain the type Fn to be a function type with any arguments.
• Then we spread the arguments of Fn and the new argument A into a new tuple. The arguments name (x) is arbitrary, we could have used any other name.
• Then we get the return type of Fn using ReturnType<Fn>.

Alternative solution without using TypeScript built-ins

type AppendArgumentNoBuiltIns<
  Fn extends (...args: any[]) => any,
  A,
> = Fn extends (...args: infer P) => infer R ? (...args: [...P, A]) => R : never

How it works:

• Same as the first solution, but without using TypeScript built-ins.
• We use a conditional type to check if Fn extends a function type with any arguments and infer the type of the arguments (P) and return type (R).
• If it does, we return a new function type with the arguments of Fn and the new argument A and also with the same return type R ((...args: [...P, A]) => R).

This challenge helps you understand TypeScript's advanced type system and how to apply this concept in real-world scenarios.