Discussion about typeof navigation

[Alphapage]

TypeScript Version: 2.5.3

Code

class C {
    constructor(s:string){}
    test(){}
}

const o={C:C}

let c:typeof o.C.prototype

let c2:C=new C('');

let c3:(typeof o).C;

Expected behavior:
(typeof o).C in c3 produces an error.
For now, typeof can navigate to variable and properties only, not types.
It would be great to be able to continue navigate types after a variable casting. So, (typeof o).C and typeof o.C.prototype will produce the same type result.
I prefer (typeof o).C of course.
Actual behavior:
let c3:(typeof o).C; hates parenthesis

[ahejlsberg]

This is already possible:

let c3: (typeof o)["C"]["prototype"];

See #11929 for more details.

[Alphapage]

You learn me something, but that doesn't change the navigation: you need to relay on variable navigation.
What I am trying to achieve is to switch from (typeof o) variable/property navigation to type navigation, so let c:C is the same as let c:(typeof o).C (this is what I call type navigation).

[ahejlsberg]

I'm afraid I don't understand what you're asking for. If T is a type, what exactly are the semantics you expect for a type T.X?

[DanielRosenwasser]

Additionally, if you want (typeof o).C to be equivalent to (typeof o)["C"], that's not possible given the way the language currently uses . for looking up a type within a namespace.

[Alphapage]

Yes, I think it could only work for object properties type or module properties type.
@ahejlsberg if T is typeof o, then T.C is C, not typeof o.C (I'm sorry for the very bad explaination).
As soon as you ask typeof o, you enter in a type navigation, so (typeof o).C is the same type as C.
@DanielRosenwasser I understand the way the language is navigating to lookup a type, but I think this could be an advanced feature to think about.

[Alphapage]

Maybe this feature is confusing for users, so I close the request.