In legacy C++ (iostream.h), what happens when a private member function is called from main versus a public one? Carefully read the code and predict the program's behavior.\n\n#include<iostream.h>\nclass CuriousTabSample\n{\n private:\n int AdditionOne(int x, int y = 1)\n {\n return x * y;\n }\n\n public:\n int AdditionTwo(int x, int y = 1)\n {\n return x / y;\n }\n};\nint main()\n{\n CuriousTabSample objCuriousTab;\n cout << objCuriousTab.AdditionOne(4, 8) << " ";\n cout << objCuriousTab.AdditionTwo(8, 8);\n return 0;\n}

Introduction / Context:
This question checks C++ access control for class members and how it affects compilation. The class exposes two member functions: one is private and multiplies two integers, while the other is public and performs integer division. The main function attempts to call both, which leads to an access-control issue at compile time.

Given Data / Assumptions:

• AdditionOne is declared in the private section of the class.
• AdditionTwo is declared in the public section.
• main calls objCuriousTab.AdditionOne(4, 8) and then objCuriousTab.AdditionTwo(8, 8).
• Legacy header is used but does not change access rules.

Concept / Approach:
In C++, class members are private by default if not specified, and any member explicitly placed under private: can be accessed only by member functions, friends, or privileged code. A call from main to a private member function is not permitted and must be rejected by the compiler.

Step-by-Step Solution:
Identify access: AdditionOne is in the private section.Attempted call: objCuriousTab.AdditionOne(4, 8) originates outside the class.C++ rule: calling a private member from outside the class is ill-formed.Therefore, compilation fails before any code runs or outputs are produced.

Verification / Alternative check:
Move AdditionOne to the public section or create a public wrapper that calls it; compilation will then succeed and you can observe outputs.

Why Other Options Are Wrong:
Option A/B/C assume successful execution. In reality, the program never links or runs due to the illegal private call.
Option E is incorrect because AdditionTwo is public and would be callable if the first error did not exist.

Common Pitfalls:
Forgetting that access control is enforced at compile time, not run time, and assuming the presence of default arguments changes accessibility (it does not).

Final Answer:
The program will report compile time error.