false error on generic type alias parameters about typescript HOT 5 CLOSED

This is a correct error; you could write

type MyRouter = {
  [Symbol.isConcatSpreadable]: number;
}
type M = GetRouteResponseInfo<MyRouter, typeof Symbol.isConcatSpreadable>;

at which point Router[Path] has type number, which there's no ReturnType<> of

from typescript.

hey @RyanCavanaugh, I'm not sure you understand my point.

TypedRouter retruns R which is contained to be at least {[key: string]: (args: any) => any;} (even if any is passed, like in TypedRouter<any>), and because Router extends TypedRouter<any>, then Router is also constrained to {[key: string]: (args: any) => any;}. so keyof Router would forever be string, and Router[Path] would forever be a valid function.

In your example, You just assumed a route key of type number which is not possible (because the constrain on R in TypedRouter). your example that not explain why this is 'Working as Intended` at all.

and if so, why does holding the type inside nested prop fixes the issue and the compiler does not complain anymore?

from typescript.

then Router is also constrained to {[key: string]: (args: any) => any;}

This index signature only constrains what's present at string keys. It doesn't bar what happens at symbol keys

type MyRouter = {
  [Symbol.isConcatSpreadable]: number;
}
type IsRouter<T extends TypedRouter<any>> = T;
type Test = IsRouter<MyRouter>; // Not a constraint violation

from typescript.

I think I get it now, and it does not relates to the way index signature checks symbols.
actually, in your example you can use 'random' literal (or any other random type), and it wouldn't be a constrain violation.

export type TypedRouter<
  R extends {
    [key: string]: (args: any) => any;
  }
> = R;

type IsRouter<T extends TypedRouter<any>> = T;

type Test = IsRouter<'random'>; // <-- Not a constraint violation

this is because when checking for T extends TypedRouter<any>, typescript checks if R=any extends {[key: string]: (args: any) => any;} (and it does), so is it returns R(any) without farther constraining it to {[key: string]: (args: any) => any;}, which we can debate if is excepted(for user perspective) or not.

in other words,

type IsRouter<T extends TypedRouter<any>> = T;

is resolved to

type IsRouter<T extends any> = T;

(because TypedRouter<any>=any)
instead of

type IsRouter<T extends {[key: string]: (args: any) => any}> = T;

(because any in TypedRouter<any> should be narrowed down to {[key: string]: (args: any) => any} as R in TypedRouter applies)

i would claim that this is not expected, and the type should be narrowed down(but that just my opinion 🫠).

nevertheless, because the type are not narrowed down, you are correct that this works as intended, and this issue can be closed.

but one last thing, is there is documentation/issue that explains the design choice to not narrowing down returned generic types by their constraint? (in other words, why TypedRouter<any>=any instead of {[key: string]: (args: any) => any})

from typescript.

This issue has been marked as "Working as Intended" and has seen no recent activity. It has been automatically closed for house-keeping purposes.

from typescript.