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Discussion on: Advanced TypeScript Exercises - Answer 6

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Replies for: So I tried in TypeScript Playground and the following seems also a valid solution for the naive problem: type NaiveFlat<T extends any[]> = ...

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Pragmatic Maciej •

Hi, thanks for the comment.
In order to avoid this message we can make a trick, and the trick is to exactly use mapped type. Pay attention that I can take your code and put into mapped type and it doesn't complain anymore:

type DeepFlat<T extends any[]> = {
  [K in keyof T] : T[number] extends any[] ? DeepFlat<T[number]> : T[number]
}[0] // 0 means we now that it will be evaluated for the first item also fully