Implement the built-in Parameters generic without using it.
For example:
const foo = (arg1: string, arg2: number): void => {}
type FunctionParamsType = MyParameters<typeof foo> // [arg1: string, arg2: number]
/* _____________ Your Code Here _____________ */
type MyParameters<T extends (...args: any[]) => any> = T extends (...args: infer Args) => any ? Args: [];
/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'
const foo = (arg1: string, arg2: number): void => {}
const bar = (arg1: boolean, arg2: { a: 'A' }): void => {}
const baz = (): void => {}
type cases = [
Expect<Equal<MyParameters<typeof foo>, [string, number]>>,
Expect<Equal<MyParameters<typeof bar>, [boolean, { a: 'A' }]>>,
Expect<Equal<MyParameters<typeof baz>, []>>,
]
标签:Typescript,const,_____________,Parameters,arg1,arg2,number,76,type From: https://www.cnblogs.com/Answer1215/p/16837966.html