题目描述
Implement a type IsUnion, which takes an input type T and returns whether T resolves to a union type.
For example:
type case1 = IsUnion<string> // false
type case2 = IsUnion<string | number> // true
type case3 = IsUnion<[string | number]> // false
题解
// ============= Test Cases =============
import type { Equal, Expect } from './test-utils'
type cases = [
Expect<Equal<IsUnion<string>, false>>,
Expect<Equal<IsUnion<string | number>, true>>,
Expect<Equal<IsUnion<'a' | 'b' | 'c' | 'd'>, true>>,
Expect<Equal<IsUnion<undefined | null | void | ''>, true>>,
Expect<Equal<IsUnion<{ a: string } | { a: number }>, true>>,
Expect<Equal<IsUnion<{ a: string | number }>, false>>,
Expect<Equal<IsUnion<[string | number]>, false>>,
// Cases where T resolves to a non-union type.
Expect<Equal<IsUnion<string | never>, false>>,
Expect<Equal<IsUnion<string | unknown>, false>>,
Expect<Equal<IsUnion<string | any>, false>>,
Expect<Equal<IsUnion<string | 'a'>, false>>,
Expect<Equal<IsUnion<never>, false>>,
]
// ============= Your Code Here =============
type IsUnion<T, U = T> =
[T extends infer X ? Exclude<U, X> : never] extends [never]
? false
: true;
引入泛型U
- 引入泛型
U，默认值为T，用于辅助判断
条件类型
type IsUnion<T, U = T> =
[T extends infer X ? Exclude<U, X> : never] extends [never]
? false
: true;
-
[T extends infer X ? Exclude<U, X> : never] extends [never]：将T的每个成员推断为X，并从U中排除X，形成一个新的联合类型 -
如果排除后的联合类型为
never，说明T不是联合类型，返回false -
否则，返回
true
