Question 1

C++ (method hiding vs base call) — what happens when a derived Display() calls Display() without qualification?  #include<iostream.h> class Base {  int x, y, z; public:  Base() { x = y = z = 0; }  Base(int xx, int yy = 'A', int zz = 'B') { x = xx; y…

Accepted Answer

Introduction / Context: This question targets name lookup and method hiding in C++. The derived class defines its own Display() that writes two values, then calls Display() again without qualification. Does it call the base or itself? Given Data / Assumptions: Derived::Display() prints x and y, then calls Display() unqualified. There is a Base::Display(), but the derived function hides it for unqualified lookup. Constructors initialize Base subobject and derived members to known values (x = 66, y = 65) before the first call. Concept / Approach: Unqualified lookup finds the nearest match in the current scope first. Since Derived declares a function named Display, an unqualified call inside Derived::Display resolves to itself, causing infinite recursion. The compiler will not implicitly resolve to Base::Display()—you must qualify it (Base::Display()) to call the base version. Step-by-Step Solution: First invocation prints 66 65 (derived members).Unqualified call then re-enters Derived::Display().The cycle repeats indefinitely, printing 66 65 until stack overflow or program termination. Verification / Alternative check: Replace the recursive call with Base::Display() to get a single line beginning with 66 65 followed by the base’s computed triple values. Why Other Options Are Wrong: Compilation error does not occur; recursion is legal. Other outputs assume a base call or single execution; here the call is recursive. Common Pitfalls: Forgetting to qualify base calls when a derived function with the same name exists; assuming overload resolution will “find” the base automatically. Final Answer: The program will produce the output 66 65 infinite number of times (or till stack memory overflow).

Question 2

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.  #include<iostream.h> class CuriousTabSample {  private:  int AdditionOne(in…

Accepted Answer

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.

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