type Push<T extends unknown[], P> = [...T, P]

Solution by ClarityOfMind #35002

type Push<T extends any[], U> = [...T, U];

Solution by raeyoung-kim #34945

type Push<T extends any[], U> = T extends never[] ? [U] : [...T, U] // your answers

Solution by 56aiden90 #34882

type Push<T extends any[], U> = [...T, U]

Solution by eunsukimme #34816

type Push<T, U> = T extends Array<unknown> ? [...T, U] :never

Solution by Yuriy113 #34735

// your answers
type Push<T extends any[], U> = [...T, U];

Solution by zeyuanHong0 #34721

type Push<T extends any[], U> = [...T,U]

Solution by lephuthuc2001 #34709

type Push<T extends readonly any[], U> = [...T, U]

Solution by Git-I985 #34705

type Push<T, U> = T extends any[] ? [...T, U] : [U]

Solution by nathan2slime #34663

type Push<T extends readonly unknown[], U> = [...T,U]

Solution by devshinthant #34549

type Push<T extends unknown[], U> = [...T, U]

Solution by binhdv155127 #34473

// 여기 풀이를 입력하세요
type Push<T extends readonly unknown[], U> = [...T, U];

Solution by LeeKangHyun #34464

type Push<T extends readonly any[], U> = [...T, U];

Solution by ktim816 #34428

// your answers
type Push<T extends any[],U> = [...T, U]

Solution by gobielJonathan #34399

type Push<T extends any[], U> = [...T, U]

Solution by bkdragon0228 #34385

type Push<T extends any[], U> = [...T, U]

Solution by rookie-luochao #34364

Just use a spread operator.

// your answers
type Push<T extends any[], U> = [...T, U]

Solution by dev-jaemin #34330

type Push<T extends any[], U> = [...T, U];

Solution by wxh-cyber #34323

type Push<Xs extends unknown[], X extends unknown> = [...Xs, X]; 

Solution by souzaramon #34270

// 你的答案

type Push<T extends unknown[], U> = [...T,U]

Solution by W-fitTiger #34249

type Push<T extends any[], U> = [...T, U]

Solution by quitone #34197

문제 설명

Array.push의 제네릭 버전을 구현하세요.

예시:

type Result = Push<[1, 2], "3">; // [1, 2, '3']

풀이

이 또한 어려운 문제는 아니었습니다. 문제중 하나인, unshift 문제를 해결했다면, 동시에 해결할 수 있는 문제였습니다.

Solution

/* _____________ 여기에 코드 입력 _____________ */

type Push<T extends any[], U> = [...T, U];

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

type cases = [
	Expect<Equal<Push<[], 1>, [1]>>,
	Expect<Equal<Push<[1, 2], "3">, [1, 2, "3"]>>,
	Expect<Equal<Push<["1", 2, "3"], boolean>, ["1", 2, "3", boolean]>>
];

Solution by adultlee #34116

type Push<T extends any[], U extends any> = [...T, U];

Solution by ProvorovOleksii #34083

type Push<T extends unknown[], U> = [...T, U];

Solution by ouzexi #33972

type Push<T extends unknown[], U> = [...T,U]

Solution by hyojinLee-git #33946

type Push<T extends unknown[], U> = [...T, U]

Solution by notsecret32 #33864

type Push<T extends any[], U> = [...T, U]

Solution by okasyun #33834

type Push<T extends any[], U> = [...T, U]

concatの問題の応用で解けた。 T型に対してextends any[]を宣言しないと...Tがエラーになる。

Solution by okabe-yuya #33826

type Push<T extends unknown[], U> = [...T, U]

Solution by veralex #33757

type Push<T extends Array<unknown>, K> = [...T, K]

Solution by Danny101201 #33739