C++ non-virtual method call through a base pointer: which class name is printed when the static type is A*? #include<iostream.h> class A { public: void CuriousTabFunction(void) { cout<< "Class A" << endl; } }; class B: public A { public: void CuriousTabFunction(void) { cout<< "Class B" << endl; } }; class C : public B { public: void CuriousTabFunction(void) { cout<< "Class C" << endl; } }; int main() { A *ptr; B objB; ptr = &objB; ptr = new C(); ptr->CuriousTabFunction(); return 0; }

Introduction / Context:
This program tests dynamic dispatch vs. static binding. Member functions are not declared virtual, so calls through a base pointer use static type information, not the dynamic type of the object.

Given Data / Assumptions:

• ptr has static type A*.
• It ultimately points to an object of type C (via new C()).
• Each class defines a method with the same name and signature, but none are virtual.

Concept / Approach:
Without virtual, the call is statically bound to A::CuriousTabFunction because the expression type is A*. Dynamic dispatch would require virtual on the base declaration to select the most-derived override at runtime.

Step-by-Step Solution:
1) ptr becomes an A* pointing at a C object. 2) ptr->CuriousTabFunction() is compiled to call A::CuriousTabFunction due to static binding. 3) The output is “Class A”.

Verification / Alternative check:
Mark CuriousTabFunction as virtual in A and rerun; the output becomes “Class C”.

Why Other Options Are Wrong:
“Class B” or “Class C” would require virtual dispatch. There is no compile-time error since method lookup succeeds and signatures match.

Common Pitfalls:
Assuming polymorphism without declaring methods virtual; forgetting that overriding without virtual still hides but does not polymorphically override.

Final Answer:
Class A.