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
.