Comments (5)
Cool project, thanks for your contribution!
BTW, I found a detailed explanation in the original type-challenges repo: type-challenges/type-challenges#614
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My two cents to enhance your great explaination. You put []
around the T
type parameter to stop the distribution over union from taking place...you don't want it in that point. You need it right after.
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You are right, thanks for clarification! Recently, I've figured it out myself by going through updated TypeScript handbook. For others to read, here is a link to how to disable distributivity over unions.
There is a small footnote:
Typically, distributivity is the desired behavior. To avoid that behavior, you can surround each side of the
extends
keyword with square brackets.
from type-challenges-solutions.
I think final solution is overcomplicated. I've used your idea (thx for project 🔥) to make up this variant. I found it more clear:
// We still need a copy of source type to have source and distributed types
type Permutation<T, C = T> =
// Here we checks that source type is not empty
[T] extends [never]
? []
// We don't care about type, just need to distribute union
: T extends unknown
// Exclude current type from source union
? [T, ...Permutation<Exclude<C, T>>]
: never
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Thanks for your contribution!
I wrote a Chinese detailed explanatation for anyone need it: https://juejin.cn/post/7165170011282079751
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