Let’s take this simplified example:

class Container<A> {
    static func foo<A>(item: A) {
        ...
    }
}

Xcode gives me a warning saying Generic parameter 'A' shadows generic parameter from outer scope with the same name; this is an error in Swift 6.

But in my particular case, shadowing the outer generic is exactly what I need to do.
The real problem comes when I want to call the method foo without specifying any generic type, so relying completely on Swift’s type inference, like:

Container.foo(item: 42)

Swift correctly infers foo’s A type as an Int, but gives me an error Generic parameter 'A' could not be inferred because it still requires Container’s A type.

Is there a way to “bypass” this and make Container not require its generic given that it has been shadowed by the foo’s one?