<@U0B8ZP13Q>: well, no. It’s `in` not because it i...

# announcements

[orangy]

@cedric: well, no. It’s in not because it is in parameter, but because it is in parameter and return value of functional type. E.g.

interface Test<out T> {
    fun test1() : () -> T  // ok, we return the lambda that will provide the value for T, so it’s `out`
    fun test2() : (T) -> Unit  // nope, we return the lambda and we will receive the value for T, so it’s `in`
}