Why in this code type `GT` cannot be inferred? ```...

# announcements

[snrostov]

Why in this code type GT cannot be inferred?

class A<T>(val t: T)
class B<T, out U: A<T>>(val t: T, val at: U)

fun <T> B<T, *>.withCapturedU(): B<T, A<T>> = this

private fun <FT> f(x: B<FT, *>) {
  g(x) // ERROR, why?
  g(x.withCapturedU()) // OK
}

private fun <GT>  g(y: B<GT, A<GT>>) {
  println(y)
}

[Andreas Sinz]

* seems to put the compiler off. if I change it to x: B<FT, A<FT> it works fine

[Andreas Sinz]

@snrostov ok, its not the *, but rather the variance of your generics, if you define T inside A and B as out, it works fine

[snrostov]

@Andreas Sinz, yes, but why first substitution in compiller error message is not suitable? Moroover it is not highlighted as red. Looks like compiller (or at least diagnostic formatting) error. And why variant using `withCpaturedU`is working?

[Andreas Sinz]

@snrostov because * can be A<FT> or any subclass of A<FT> together with FT being invariant means the compiler is not able to ensure that you'll always have A<GT> inside your B<GT, *>. not sure about the red highlighting

[snrostov]

Ah, I see now. Thanks for explanation!