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

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

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

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

This?

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

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"