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Comments (5)

nmonastyrskyi avatar nmonastyrskyi commented on May 24, 2024 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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jfet97 avatar jfet97 commented on May 24, 2024 3

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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ghaiklor avatar ghaiklor commented on May 24, 2024 3

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.

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ilmpc avatar ilmpc commented on May 24, 2024 2

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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zhaoyao91 avatar zhaoyao91 commented on May 24, 2024 1

Thanks for your contribution!
I wrote a Chinese detailed explanatation for anyone need it: https://juejin.cn/post/7165170011282079751

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