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I am working on typescript-challenge 399. Tuple Filter (https://github.com/type-challenges/type-challenges/blob/main/questions/00399-hard-tuple-filter/README.md)

This is my code

type FilterOut<T extends any[], F, Res extends unknown[] = []> = 
  T extends [infer X, ...infer Y]
  ? X extends F
    ? FilterOut<Y, F, Res>
    : FilterOut<Y, F, [...Res, X]>
  : Res

type t0 = FilterOut<['a', 'b'],'a'> // ['b'] -> Just as I expected
type t1 = FilterOut<[never], never> // never -> But I expect []
type t2 = FilterOut<[never, 'a'], never> // never -> But I expect ['a']

As you can see, t1 and t2 are different from my expectation. I know if I replace X extends F to [X] extends [F], then I can fix this question.

However, I just wonder why we will obtain never in the first version code? Is this due to some weird properties of never?

Here I try another thing about never, and I think it maybe due to the same reason from the previous question.

type t3 = never extends never ? 1 : 2 // 1 -> As I expected
type test<T, F> =
  T extends F 
    ? 1
    : 2 

type t4 = test<never, never> // never -> But I expect 1

Can anybody help me or give me some documentation about this question! Thanks a lot!

kelsny
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Jiangqi Zhu
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  • `test` is a *distributive* conditional type, and `never` is considered the empty union, so it will output `never` no matter what. `t3` is not distributive over unions. See the linked questions and answers for details and more information. – jcalz Apr 06 '23 at 17:42

0 Answers0