Typescript - add one argument to a function's params

Question

type SomeFunc = (a:string, b:number, c:someCustomType) => number;

I want to create a type that is just like the one above, except there is a single parameter added at the end. Let's say, d:number;

type SomeFuncAltered = (a:string, b:number, c:someCustomType, d:number) => number;

I do not want to craft the entire type manually though, I'm pretty sure there's a smart trick with Parameters<func> to be used here.

Answer 1

As the accepted answer here doesn't seem to work for me (using TS 4), I wrote my own type that adds any number of parameters to the end of a function.

type AddParameters<
  TFunction extends (...args: any) => any,
  TParameters extends [...args: any]
> = (
  ...args: [...Parameters<TFunction>, ...TParameters]
) => ReturnType<TFunction>;

Usage:

type SomeFunc = (a: string, b: number, c: someCustomType) => number;
type SomeFuncAltered = AddParameters<SomeFunc, [d: number]>;
// SomeFuncAltered = (a:string, b:number, c:someCustomType, d:number) => number;

Note: This solution also allows you to add named parameters.