Is there ArgumentsType like ReturnType in Typescript?

Question

ReturnType extracts return type of a function.

Is there a way to define ArgumentsType that extracts parameter types of a

Answer 1

There is no way for now to extract both the types and amount of arguments for any function possible. But you can try something like this:

type ArgumentTypes = T extends () => any ? never[] :
                        T extends (a1: infer T1) => any ? [T1] :
                        T extends (a1: infer T1, a2: infer T2) => any ? [T1, T2] :
                        // continue here for any reasonable number of args
                        never;

Check it with the following:

const args0: ArgumentTypes<() => boolean> = []; // correct
const args1: ArgumentTypes<(a: number) => boolean> = [1]; // correct
const args2: ArgumentTypes<(a: number, b: string) => boolean> = [1, 'str']; // correct

const oops0null: ArgumentTypes<() => boolean> = null; // error, arguments are array (but empty one)
const oops01: ArgumentTypes<() => boolean> = [1]; // error, arguments must be empty

const oops10: ArgumentTypes<(a: number) => boolean> = []; // error, we need one argument
const oops12: ArgumentTypes<(a: number) => boolean> = [1, 2]; // error, we need only one argument
const oops1wrong: ArgumentTypes<(a: number) => boolean> = ['str']; // error, argument must be number

const oops21: ArgumentTypes<(a: number, b: string) => boolean> = [1]; // error, we need two arguments
const oops23: ArgumentTypes<(a: number, b: string) => boolean> = [1, 'str', undefined]; // error, we need only two arguments
const oops2wrong: ArgumentTypes<(a: number, b: string) => boolean> = ['str', 1]; // error, arguments are reversed

Note that this don't have any use of optional arguments - they are just omitted from the output. I wasn't able to find a way to catch them for now.