Why does `foo.value` resolve to `Any?` and not the upper bound `Bar`?

# announcements

[Marc Knaup]

Why does foo.value resolve to Any? and not the upper bound Bar?

[thana]

in makes the type parameter contravariant. you can safeley write a Bar to foo but can't safley read it

[Marc Knaup]

I know, but even though it’s contravariant it can’t be higher in hierarchy than Bar, which is the upper bound.

[turansky]

https://youtrack.jetbrains.com/issue/KT-38134

This?

[Marc Knaup]

Thank you, but I think that’s a different once.
Mine is about contravariance, also it also doesn’t work with old inference.

[thana]

i still think it's working as expected. of course you restricted T to be a Bar but still in Bar mean "any supertype of Bar"