C++ operator overloading with a default argument on operator+ (member): compute the result of objA + 5 and print it. #include <iostream.h> class Addition { int x; public: Addition() { x = 0; } Addition(int xx) { x = xx; } Addition operator+(int xx = 0) { Addition objTemp; objTemp.x = x + xx; return objTemp; } void Display() { cout << x << endl; } }; int main() { Addition objA(15), objB; objB = objA + 5; objB.Display(); return 0; }

Introduction / Context:
This question validates understanding of member operator overloading and default arguments. A member operator+(int) computes a new Addition whose x is the receiver’s x plus the provided integer. Default arguments on operators are allowed in C++ and are not the cause of any compilation issue here.

Given Data / Assumptions:

• objA is constructed with x = 15.
• Expression objA + 5 invokes Addition::operator+(int).
• The operator returns a new Addition with x = 15 + 5 = 20, then Display prints that value.

Concept / Approach:
Member operator+(int) is analogous to a method call objA.operator+(5). It builds a temporary result object with the sum and returns it by value. No undefined behavior occurs; the temporary is copied into objB.

Step-by-Step Solution:

Verification / Alternative check:
If you omit the argument (objA + ), the default argument 0 would be used and the result would be 15. The code as written provides 5 explicitly.

Why Other Options Are Wrong:

• Run time error / garbage: nothing invalid or uninitialized occurs.
• Compilation fails due to operator + cannot have default arguments.: False; operators may have default parameters like any function, subject to signature rules.

Common Pitfalls:
Confusing member versus non-member operator overload constraints, or assuming default arguments are disallowed for overloaded operators.

Final Answer:
The program will print the output 20.