On Wednesday, 12 March 2014 at 13:05:07 UTC, Steve Teale wrote:
I a = new A();Seems like a bug to me.
You are working through the interface, so the type will be what was written in the interface. If you made that "A a = new A();", then you could see type A on the override.
But the interface has no way of knowing which child class is there at compile time so it has to work through base types only.
