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 = x + yy; z = x + y; } void Display(void) { cout << x << " " << y << " " << z << endl; } }; class Derived : public Base { int x, y; public: Derived(int xx = 65, int yy = 66) : Base(xx, yy) { y = xx; x = yy; } void Display(void) { cout << x << " " << y << " "; Display(); } }; int main() { Derived objD; objD.Display(); return 0; } Choose the correct statement.

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:

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).