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
Not fully understand how the solution works.
/* _____________ Your Code Here _____________ */
type IsUnion<T, C = T> = [T] extends [never]
? false
: T extends C
? [C] extends [T]
? false
: true
: never;
// [T]: [void | "" | null | undefined]
// [C]: [void] | [""] | [null] | [undefined]
type x= IsUnion<undefined | null | void | ''>
/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/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>>,
]标签：IsUnion,Typescript,false,_____________,true,Medium,Expect,type From： https://www.cnblogs.com/Answer1215/p/16743629.html