Hi, i have some class like `class Foo<T> (v...

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[thana]

Hi, i have some class like class Foo<T> (val helper1: Helper1<T>, helper2: Helper2<T>) . here i want Helper1 and Helper2 work on the same type T but in fact Foo doesn't really care about what T actually is - they just must match.
Is there a way not having to declare the T on Foo itself? I don't want to write something like Foo<*> everywhere i'm using Foo
Or maybe the better question: i sense there is some design flaw here but i cannot really put the finger on. What approaches are there, to resolve this flaw?

[E.Kisaragi]

don't you want Foo to have generic type T, right?

[thana]

Foo shouldn't be generic

[E.Kisaragi]

you may try this:
class Foo {
companion object {
operator fun <T> invoke(help1: Helper1<T>,help2: Helper2<T>): Foo = TODO("logic")

}
}

[E.Kisaragi]

and Foo(helper1, helper2)

[thana]

but then, how do i declare the fields helper1 and helper2 ?

[dead.fish]

The invoke operator above acts as factory, you add them to some private constructor where your args become Helper<Any>

[dead.fish]

after all, Foo shouldn’t care about the `Helper`s generic args, right?

[thana]

ah good point

[thana]

i dpo not realy like the Any approach, but at least it's a small improvement... i hope 😉
Thank you so far!

[thana]

still i feel there is something wrong with that approach as a whole. any ideads on that?

[Luis Mirabal]

A little bit more context would help. I t depends what role do the helpers play in Foo. If they’re dependencies and Foo can’t perform its responsibilities without them, it would make sense that Foo includes the generic type. If the helpers play a secondary role, maybe a type wrapping them and serving as a facade make sense, and then that new type can include the generic info