Typescript：创建通用展开<t>辅助类型</t>

Question

I've created a helper type to unwrap the inner type of an Observable<T> :

type UnwrapObservable<T> = T extends Observable<(infer U)> ? U : never;

This works just fine. But... my next thought was, "How can I make this type completely generic and apply to any type, not just Observable, essentially Unwrap<T> ?

Here's what I tried:

type Unwrap<T, W> = T extends W<(infer U)> ? U : never;

But I get:

Type 'W' is not generic. ts(2315)

How can I create a type that unwraps any specified type?

Answer 1

What You are trying to do is impossible in TS, but as far as I know it is possible in Flow)

In TS you can't apply restrictions for generic which expects other generic as argument.

This is the most generic Wrap type

type Wrapp<T> = { wrapped: T }
type Unwrap<T> = T extends Wrapp<(infer R)> ? R : never

Please see here for similar limitation