Answer
This question is compilable and deterministic.
Its output is “A”.
Explanation
As long as A::f() is not virtual, A::f() will always be called, even if the reference or pointer is actually referring to an object of type B.
This question is compilable and deterministic.
Its output is “A”.
As long as A::f() is not virtual, A::f() will always be called, even if the reference or pointer is actually referring to an object of type B.



