Conditional types infer gets wrong type if it inferred from union type

[ktsn]

TypeScript Version: 3.9.0-dev.20200330

Search Terms: infer function union conditional types

Code

interface A {
  test: never
}

type PropType<T> = 
  | { new(...args: any[]): T & object }
  | { new(...args: string[]): Function }

type ExtractProp<P extends PropType<any>> = P extends PropType<infer T> ? T : never

type Res = ExtractProp<PropType<A>> // Function | A

Expected behavior:
Res is of type A

Actual behavior:
Res is of type Function | A

Playground Link:
https://www.typescriptlang.org/play/?ssl=12&ssc=1&pln=1&pc=1#code/JYOwLgpgTgZghgYwgAgILIN4FgBQzmQDOYAXMiBAG7S4C+uuYAngA4oAKUA9iwCqsQAPLwB8yALzJc+AD6ZyEAO4AKAHTq4UAOaEycEEwDaAXQCUZXsgBkyLgCMAVhARhk9PMjkYFK9as06ZMRQoFom5sgAYgCuIC7AXCBuDDjMbMgAogAeYFCIYJw8guzIEDkQIAAmhMiFfAKC+kwiYpIlZZBVNXX8bIKgMNDIosgA-MPIZBTUUClpKABKEDWS2bn5dcXc9X2oLcgA9AdRsfGJnmi4QA

Related Issues:
vuejs/composition-api#291

Background:
In Vue.js, we annotate props by using built in PropType. Since PropType accepts any constructors, we have all possible cases in it as union type. (e.g. new(...args: any[]): T & object is for user defined class constructor, new(...args: string[]): Function is for Function constructor)

To use the props type, we need to extract actual type from PropType generics (ExtractProp in reproduction). We want to extract type A if the constructor type is of PropType<A>. But the actual behavior is Function | A.

I also found if I pass PropType via another object type, it infers correctly.
https://www.typescriptlang.org/play/#code/JYOwLgpgTgZghgYwgAgILIN4FgBQzmQDOYAXMiBAG7S4C+uuYAngA4oAKUA9iwCqsQAPLwB8yALzJc+AD6ZyEAO4AKAHTq4UAOaEycEEwDaAXQCUZXsgBkyLgCMAVhARhk9PMjkYFK9as06ZMRQoFom5sgAYgCuIC7AXCBuDDjMbMgAogAeYFCIYJw8guxikuzIEDkQIAAmhPJpEGSFfAKCoDDQyKJuyAD83chkFNRQKY3IAEoQ9ZLZufktgt6NzdytbIKoYrRiAPR7aLhAA

[RyanCavanaugh]

It sounds like you don't want this conditional type to be distributive, so you should write

type ExtractProp<P extends PropType<any>> = [P] extends [PropType<infer T>] ? T : never

[ktsn]

Ah, I forgot about distribution of conditional types. Thank you for providing the code!

Although the original issue is solved by avoiding distribution, I'm still not sure why Function comes in the result type as the inferred type parameter T is not relevant type of Function and Function part of PropType does not include type parameter T. Would you mind telling me how type inference works under the hood in this case?

[typescript-bot]

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