Override TypeScript interface with multiple generics

Issue

I have the following scenario:

interface Foo<GenericA, GenericB> {}

I want to declare a variable of type Foo, but only specify the second generic, e.g.:

const foo: Foo<GenericBType>;

I can only do that if I write

const foo: Foo<unknown, GenericBType>;

It is not really that bad, but it doesn't feel right either, as I just want to make use of the second type, and I am not interested in the first at all.

Is there any other way I can write this in a more succinct way?

PS: This is about the expressjs types. I want to type the Request object, but I only want to consider the type of the Query params, which is the fourth generic in the type of the Request:

 interface Request<
    P = core.ParamsDictionary,
    ResBody = any,
    ReqBody = any,
    ReqQuery = core.Query,
    Locals extends Record<string, any> = Record<string, any>
> extends core.Request<P, ResBody, ReqBody, ReqQuery, Locals> {}

so I would have to write something like:

req: Request<unknown, unknown, unknown, Foo>

and it's really not beautiful...

Solution

The simplest solution here is to create type alias:

interface Foo<GenericA, GenericB> { }

type Foo2<B> = Foo<unknown, B>

declare const foo: Foo2<'generic B'>

If all generics are optional then if you provide only one generic argument TS compiler will treat it as a first generic. There is no wildcard like _ or *. I think using unknown is the best way to handle missing generics.

You might looking for existential types (issues/10571). This is not supported by typescript. Existential types feature was supported in Flow but as you might have noticed it is deprecated now due to unsafety:

Triggers when you use the * (existential) type, as this type is unsafe and usually just equivalent to any. The effect of * can generally be achieved by simply not providing a type annotation.

Answered By - captain-yossarian