type MyEqual<X, Y> =
  (<T>() => T extends X ? 1 : 2) extends
  (<T>() => T extends Y ? 1 : 2) ? true : false
type Includes<T extends readonly any[], U> = T extends [] ? false :
  MyEqual<T[0], U> extends true
  ? true
  : T extends [infer f, ...infer rest]
  ? Includes<rest, U>
  : false;

Solution by watanaki #35195

type IsEquar<X, Y> = (<T>() => T extends X ? 1 : 2) extends <T>() => T extends Y ? 1 : 2 ? true : false;
type Includes<T extends readonly any[], U> = T extends [infer F, ...infer Rest]
  ? IsEquar<F, U> extends true
    ? true
    : Includes<Rest, U>
  : false;

Solution by wendao-liu #35085

My answer is inspired by util implementation of Equal but more understandable i think. It passes all given tests

type MyEquals<T,U> = 
  Readonly<T> extends T ? Readonly<U> extends T ? true : false : false;
  
type Includes<T extends readonly any[], U> = 
  T extends readonly [infer HEAD, ...infer TAIL] ? 
  Equal<HEAD,U> extends true ? true : Includes<TAIL, U> : 
    false;

Solution by CrimsoonXIII #35047

type Includes<T extends readonly unknown[], P> = P extends T[number] ? true : false

Solution by ClarityOfMind #34994

type Includes<T extends readonly any[], U> = T extends [infer First, ...infer Rest] ? (First extends U ? (U extends First ? true : false) : Includes<Rest, U>) : false;

Solution by raeyoung-kim #34957

type TupleToUnion<T extends readonly any[]> = T extends [infer First, ...infer Rest] ? First | TupleToUnion<Rest> : never;
type Includes<T extends any[], K> = K extends TupleToUnion<T> ? true : false;

We could make a TupleToUnion.

Solution by DearICE #34907

type Includes<T extends readonly any[], U> =
  Equal<U, T[0]> extends true 
  ? true
  : (T extends [any, ...infer R] 
    ? (R extends [] // if R is an empty array, the search should be terminated 
      ? false 
      : Includes<R, U>) 
    : false)

Solution by floatDreamWithSong #34871

type Includes<T extends readonly any[], K> = K extends T[number] ? true : false

Solution by weitongtong #34847

type TupleIndexes<T extends readonly any[]> = { [K in keyof T]: K }[number] 获取元组T的所有索引 type AllEqual<T extends readonly any[], U, P extends number> = P extends P ? Equal<T[P], U> : never 利用Distributive Conditional Type，如果T中所有元素都和U不相等则为false。若T恰有一个元素和U相等则为true，若T含有元素和U相等则为boolean。 type Includes<T extends readonly any[], U> = AllEqual<T, U, TupleIndexes<T>> extends false ? false : true

Solution by 2083335157 #34838

// @reference https://github.com/microsoft/TypeScript/issues/27024#issuecomment-421529650
type IsEqual<X, Y> = 
  (<T>() => T extends X ? 1 : 2) extends 
  (<T>() => T extends Y ? 1 : 2) 
    ? true
    : false;

type Includes<T extends readonly any[], U> = T extends [infer First, ...infer Rest] 
  ? IsEqual<First, U> extends true 
    ? true 
    : Includes<Rest, U> 
  : false;

Solution by eunsukimme #34815

// 你的答案
type Includes<T extends readonly any[], U> = {
  [key in keyof T]: Equal<T[key], U> extends true ? true: never;
}[number] extends never ? false : true
// 借助 never 空集

Solution by Atlas-lili #34729

// your answers
type Includes<T extends readonly any[], U> = T extends [infer P, ...infer R]
  ? Equal<P, U> extends true
    ? true
    : Includes<R, U>
  : false;

Solution by zeyuanHong0 #34719

type Includes<T extends readonly any[], U> = T extends [infer I, ...infer E] ? Equal<I, U> extends true ? true : Includes<E, U> : false

Solution by nathan2slime #34662

type Includes<T extends readonly unknown[], U> =
  T extends [infer First, ...infer Rest]
    ? Equal<First, U> extends true ? true : Includes<Rest, U>
    : false;

Solution by devshinthant #34548

type Includes<T extends readonly any[], U> = 
T extends [infer Head, ... infer Tail] ? 
Equal<Head, U> extends true ? true : Includes<Tail, U>
: false

Solution by binhdv155127 #34470

// 여기 풀이를 입력하세요
type Equal<X, Y> =
  (<T>() => T extends X ? 1 : 2) extends (<T>() => T extends Y ? 1 : 2) ? true : false;

type Includes<T extends readonly any[], U> = T extends [infer F, ...infer Rest]
  ? Equal<F, U> extends true
    ? true
    : Includes<Rest, U>
  : false;

Solution by LeeKangHyun #34463

type Equal<X, Y> =
(<T>() => T extends X ? 1 : 2) extends
(<T>() => T extends Y ? 1 : 2) ? true : false;
type Includes<T extends readonly any[], U> = T extends [infer F, ...infer R] ? Equal<F, U> extends true ? true : Includes<R, U> : false

Solution by ktim816 #34427

type Includes<T extends readonly any[], U> = 
  T extends [infer F, ...infer R] 
    ? Equal<F, U> extends true
      ? true 
      : Includes<R, U> 
    : false;

Solution by bkdragon0228 #34377

type Includes<T extends readonly any[], U> = T extends [infer Fisrt, ...infer Reset]
  ? Equal<Fisrt, U> extends true ? true : Includes<Reset, U>
  : false

Solution by rookie-luochao #34363

export type IsEqual<X, Y> =
  (<T>() => T extends X ? 1 : 2) extends
  (<T>() => T extends Y ? 1 : 2) ? true : false

type Includes<T extends readonly any[], U> = T extends [infer A,...infer Reset] ? IsEqual <U,A> extends true ?true:Includes<Reset,U> : false

Solution by FrankWangMing #34340

type Eq<X, Y> = (<T>() => T extends X ? 1 : 2) extends <T>() => T extends Y ? 1 : 2 ? true : false

type Includes<T extends readonly any[], U> = 
  T extends [infer Head, ...infer Tail]
    ? Eq<U, Head> extends true 
      ? true
      : Includes<Tail, U>
    : false

Solution by souzaramon #34335

문제 설명

JavaScript의 Array.includes 함수를 타입 시스템에서 구현하세요. 타입은 두 인수를 받고, true 또는 false를 반환해야 합니다.

예시:

type isPillarMen = Includes<["Kars", "Esidisi", "Wamuu", "Santana"], "Dio">; // expected to be `false`

풀이

보자마자 시도한 방법은 다음과 같습니다.

type Includes<T extends readonly any[], U> = U extends T[number] ? true : false;

하지만 다음과 같은 에러가 발생했습니다.

Includes<[{}], { a: "A" }>; // true...?

해당 타입이 true 가 나오는것이 이상했습니다. 그 이유는 {} 타입은 "빈 객체 타입"으로, 모든 객체 타입의 상위 타입(supertype)으로 간주되기 때문입니다.

{} 타입의 의미

TypeScript에서 {} 타입은 **"빈 객체 타입"**으로 알려져 있습니다. 이는 객체를 의미하지만, 그 객체에 어떤 특정한 속성이 있음을 명시하지 않은 타입입니다.

{} 타입은 사실상 모든 JavaScript 값과 호환됩니다. 이는 숫자, 문자열, 배열, 함수, 객체, null, undefined를 포함한 모든 값이 {} 타입에 할당될 수 있음을 의미합니다.

let emptyObject: {} = {}; // 빈 객체 할당 가능
let numberValue: {} = 123; // 숫자 할당 가능
let stringValue: {} = "Hello"; // 문자열 할당 가능
let arrayValue: {} = [1, 2, 3]; // 배열 할당 가능
let functionValue: {} = () => {}; // 함수 할당 가능

// 더 구체적인 객체 타입에도 할당 가능
let specificObject: {} = { a: "hello", b: 42 };

object 타입은 원시 값(primitive types)을 허용하지 않지만, {} 타입은 원시 값도 허용합니다. 예를 들어, number나 string도 {} 타입에 할당될 수 있지만, object 타입에는 할당될 수 없습니다.

따라서 {} 내부에서도 확인하여 완벽히 호환하는지 확인할 필요가 있었습니다.

제귀적 풀이

그렇다면 제귀적인 방식으로 이런건 어떨까요? 저희가 DeepReadonly 를 만들어서 제귀적으로 문제를 해결했던것처럼 (저의 readonly에 풀이가 적혀 있습니다.) 이번에도 제귀적으로 문제를 풀어보겠습니다.

type Includes<Value extends any[], Item> = Value extends [
	infer First,
	...infer Rest
]
	? Equal<First, Item> extends true // Equal을 사용해보았습니다! 당연히 사용하면 안되겠지만 말이죠.
		? true
		: Includes<Rest, Item>
	: false;

/* _____________ 테스트 케이스 _____________ */
import type { Equal, Expect } from "@type-challenges/utils";

type cases = [
	Expect<
		Equal<Includes<["Kars", "Esidisi", "Wamuu", "Santana"], "Kars">, true>
	>,
	Expect<
		Equal<Includes<["Kars", "Esidisi", "Wamuu", "Santana"], "Dio">, false>
	>,
	Expect<Equal<Includes<[1, 2, 3, 5, 6, 7], 7>, true>>,
	Expect<Equal<Includes<[1, 2, 3, 5, 6, 7], 4>, false>>,
	Expect<Equal<Includes<[1, 2, 3], 2>, true>>,
	Expect<Equal<Includes<[1, 2, 3], 1>, true>>,
	Expect<Equal<Includes<[{}], { a: "A" }>, false>>,
	Expect<Equal<Includes<[boolean, 2, 3, 5, 6, 7], false>, false>>,
	Expect<Equal<Includes<[true, 2, 3, 5, 6, 7], boolean>, false>>,
	Expect<Equal<Includes<[false, 2, 3, 5, 6, 7], false>, true>>,
	Expect<Equal<Includes<[{ a: "A" }], { readonly a: "A" }>, false>>,
	Expect<Equal<Includes<[{ readonly a: "A" }], { a: "A" }>, false>>,
	Expect<Equal<Includes<[1], 1 | 2>, false>>,
	Expect<Equal<Includes<[1 | 2], 1>, false>>,
	Expect<Equal<Includes<[null], undefined>, false>>,
	Expect<Equal<Includes<[undefined], null>, false>>
];

문제없이 모두 잘 반환하는것을 확인할 수 있었습니다. 그렇다면 이제 Equal을 대신할 수 있는 함수를 만들면 되겠군요. 안타깝게도 Equal을 대신하는 제네릭은 없으니 직접 구현을 통해 만들어 주어야 합니다.

/**
Returns a boolean for whether given two types are equal.
@link https://github.com/microsoft/TypeScript/issues/27024#issuecomment-421529650
*/
type IsEqual<T, U> = (<G>() => G extends T ? 1 : 2) extends <G>() => G extends U
	? 1
	: 2
	? true
	: false;

다음과 같이 ts issues에 포함되어 있는 IsEqual을 사용해보았습니다.

Solution

/* _____________ 여기에 코드 입력 _____________ */

type IsEqual<T, U> = (<G>() => G extends T ? 1 : 2) extends <G>() => G extends U
	? 1
	: 2
	? true
	: false;

type Includes<Value extends any[], Item> = Value extends [
	infer First,
	...infer Rest
]
	? IsEqual<First, Item> extends true
		? true
		: Includes<Rest, Item>
	: false;

/* _____________ 테스트 케이스 _____________ */
import type { Equal, Expect } from "@type-challenges/utils";

type cases = [
	Expect<
		Equal<Includes<["Kars", "Esidisi", "Wamuu", "Santana"], "Kars">, true>
	>,
	Expect<
		Equal<Includes<["Kars", "Esidisi", "Wamuu", "Santana"], "Dio">, false>
	>,
	Expect<Equal<Includes<[1, 2, 3, 5, 6, 7], 7>, true>>,
	Expect<Equal<Includes<[1, 2, 3, 5, 6, 7], 4>, false>>,
	Expect<Equal<Includes<[1, 2, 3], 2>, true>>,
	Expect<Equal<Includes<[1, 2, 3], 1>, true>>,
	Expect<Equal<Includes<[{}], { a: "A" }>, false>>,
	Expect<Equal<Includes<[boolean, 2, 3, 5, 6, 7], false>, false>>,
	Expect<Equal<Includes<[true, 2, 3, 5, 6, 7], boolean>, false>>,
	Expect<Equal<Includes<[false, 2, 3, 5, 6, 7], false>, true>>,
	Expect<Equal<Includes<[{ a: "A" }], { readonly a: "A" }>, false>>,
	Expect<Equal<Includes<[{ readonly a: "A" }], { a: "A" }>, false>>,
	Expect<Equal<Includes<[1], 1 | 2>, false>>,
	Expect<Equal<Includes<[1 | 2], 1>, false>>,
	Expect<Equal<Includes<[null], undefined>, false>>,
	Expect<Equal<Includes<[undefined], null>, false>>
];

Solution by adultlee #34121

type Includes<T extends readonly any[], U> = T extends [infer P, ...infer rest] ? Equal<P, U> extends true ? true : Includes<rest, U> : false

Solution by ouzexi #33971

type Includes<T extends readonly any[], U> = T extends [infer First, ...infer Rest] 
  ? Equal<U, First> extends true
    ? true
    : Includes<Rest, U>
  : false

Solution by notsecret32 #33863

type Includes<T extends readonly any[], U> =  T extends [infer L, ...infer R] 
  ? Equal<L, U> extends true
    ? true
    : Includes<R, U>
  : false

Solution by veralex #33756

type Equal<A, B> = (<T>() => T extends A ? 1 : 0) extends (<T>() => T extends B ? 1 : 0) ? true : false

type Includes<T extends readonly any[], U> = T extends [infer A, ...infer Rest]
  ? Equal<A, U> extends true
    ? true
    : Includes<Rest, U>
  : false

Solution by laplace1009 #33707

type Includes<T extends readonly any[], U> = 
  T extends [infer Head , ...infer Tail]
    ? Equal<Head, U> extends true
      ? true
      : Includes<Tail, U>
    : false;

難しすぎる... Head, Tailがここで出てくるとは思わなかった。 Equal使うのは想像できなかった

Solution by okabe-yuya #33687

// 你的答案
type Includes<T extends unknown[], K> = K extends keyof T[number] ? true : false;

Solution by SZGG2333 #33602

type Includes<P extends readonly any[], U> = true extends {
      [K in keyof P]: (<T>() => T extends U ? 1 : 2) extends <T>() => T extends P[K] ? 1 : 2 
        ? true 
        : false;
    }[number] 
  ? true : false;

Solution by Alex-Nicalace #33406

// 여기 풀이를 입력하세요
type Includes<T extends readonly any[], U> = T extends [infer F, ...infer Rest]
  ? Equal<F, U> extends true
    ? true
    : Includes<Rest, U>
  : false;

Solution by awesomelon #33368