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Chainable Options

提出詳細

type Without<T, K> = K extends keyof T ? never : K

type Chainable<C extends { [k: number | string]: unknown } = {}> = {
  option<K extends string, V>(key: Without<C, K>, value: V):
    K extends keyof C ? Chainable<Omit<C, K> & { [P in K]: V }> : Chainable<C & { [P in K]: V }>
  get(): C
}
提出日時2023-08-08 15:56:08
問題Chainable Options
ユーザーookkoouu
ステータスAccepted
テストケース
import type { Alike, Expect } from '@type-challenges/utils'

declare const a: Chainable

const result1 = a
  .option('foo', 123)
  .option('bar', { value: 'Hello World' })
  .option('name', 'type-challenges')
  .get()

const result2 = a
  .option('name', 'another name')
  // @ts-expect-error
  .option('name', 'last name')
  .get()

type cases = [
  Expect<Alike<typeof result1, Expected1>>,
  Expect<Alike<typeof result2, Expected2>>,
]

type Expected1 = {
  foo: number
  bar: {
    value: string
  }
  name: string
}

type Expected2 = {
  name: string
}