C++ (name qualification in multiple-base-like contexts) — evaluate the printed sum after calling the specifically qualified BaseTwo::Display. #include static double gDouble; static float gFloat; static double gChar; static double gSum = 0; class BaseOne { public: void Display(double x = 0.0, float y = 0.0, char z = 'A') { gDouble = x; gFloat = y; gChar = int(z); gSum = gDouble + gFloat + gChar; cout << gSum; } }; class BaseTwo { public: void Display(int x = 1, float y = 0.0, char z = 'A') { gDouble = x; gFloat = y; gChar = int(z); gSum = gDouble + gFloat + gChar; cout << gSum; } }; class Derived : public BaseOne, public BaseTwo { void Show() { cout << gSum; } }; int main() { Derived objDev; objDev.BaseTwo::Display(10, 20, 'Z'); return 0; } What value is printed?

Introduction / Context:This question reinforces qualified calls to resolve ambiguity in multiply-inherited interfaces and shows how arguments flow into a function with defaults overridden at the call-site. The output is a straightforward arithmetic sum of three values stored in globals.

Given Data / Assumptions:

• We explicitly call objDev.BaseTwo::Display(10, 20, 'Z').
• Inside BaseTwo::Display, the globals are set and gSum is computed as the sum of the three converted values.
• 'Z' has integer code 90.

Concept / Approach:The computation uses gDouble = 10, gFloat = 20.0f, gChar = 90, so the sum is 10 + 20 + 90. The function itself streams that sum via cout, so the visible output is the numeric result.

Step-by-Step Solution:

Verification / Alternative check:Calling BaseOne::Display(10.0, 20.0f, 'Z') would compute the same numeric sum since the same values are used after conversions.

Why Other Options Are Wrong:

• 0 / garbage: contradicts the explicit arithmetic.
• Errors: The qualified call resolves any ambiguity; the code compiles and runs.

Common Pitfalls:Forgetting to qualify the call in multiple inheritance, which would make the call ambiguous and fail to compile.

Final Answer:The program will print the output 120.