C++ (inheritance, default arguments, and overrides) — track member initialization and outputs with pre/post operators. #include<iostream.h> class CuriousTabBase { public: int x, y; CuriousTabBase(int xx = 0, int yy = 5) { x = ++xx; y = --yy; } void Display() { cout << --y; } ~CuriousTabBase() {} }; class CuriousTabDerived : public CuriousTabBase { public: void Increment() { y++; } void Display() { cout << --y; } }; int main(){ CuriousTabDerived objCuriousTab; // uses base default ctor objCuriousTab.Increment(); objCuriousTab.Display(); return 0; } What is printed?

Introduction / Context:This question tests default argument handling in a base constructor and how pre/post operators affect state across member functions, including an override of Display() in the derived class.

Given Data / Assumptions:

• Base constructor runs with defaults xx = 0, yy = 5.
• In constructor: x = ++xx → 1; y = --yy → 4.
• Derived adds Increment() → y++.
• Derived overrides Display() to print --y.

Concept / Approach:Trace the member y across operations in order: initialization, increment, and pre-decrement during display. The code never uses the base Display(); it uses the override in the derived type, operating on the same y member inherited from the base.

Step-by-Step Solution:

Verification / Alternative check:If you instead called the base Display() (e.g., objCuriousTab.CuriousTabBase::Display()), the same calculation would occur since both versions print --y.

Why Other Options Are Wrong:

• 3 or 5: Off-by-one due to misreading pre/post semantics.
• Garbage/compile error: All members are initialized and types valid.

Common Pitfalls:Forgetting that ++ and -- change the stored value, not just the printed value.

Final Answer:4