Distributive Conditional Types & Mapped Types does not work together

[jiamingparker]

For challenge 20, the most rated solution does not work for test case 4 for me:

declare function PromiseAll<T extends any[]>(values: readonly [...T]): Promise<{ [K in keyof T]: T[K] extends Promise<infer R> ? R : T[K] }>;

I have to move the extends part to another generic type

type MyAwaited<T> = T extends Promise<infer R> ? R : T
declare function PromiseAll<T extends any[]>(values: readonly [...T]): Promise<{ [K in keyof T]: MyAwaited<T[K]> }>;

This is really confusing. My explanation is distributive conditional types does not work in mapped types. When T[K] is a union type number | Promise<number>, the distributive behavior does not happen. But if we use another generic type, the context is clear and the distributive behavior happen again.

Anybody can explain on this?

[Zilong Yao]

For the first case:

declare function PromiseAll<T extends any[]>(values: readonly [...T]): Promise<{ [K in keyof T]: T[K] extends Promise<infer R> ? R : T[K] }>;

It equals to:

// replace T[K] to number | Promise<number>
type Value = number | Promise<number> extends Promise<infer R> ? R : number | Promise<number>;

It obviously will return number | Promise<number>.

But in the second case, you abstract the judge Statement, so it will be:

type MyAwaited<T> = T extends Promise<infer R> ? R : T;
type Value = MyAwaited<number | Promise<number>> = MyAwaited<number> |  MyAwaited<Promise<number>> = number;

So it works fine.