In C++ (iostream.h), a class method with default parameters modifies members and prints an expression using pre-increment and pre-decrement. Determine the output.\n\n#include <iostream.h>\n\nvoid Tester(int xx, int yy = 5); // unrelated free function\n\nclass CuriousTab\n{\n int x;\n int y;\npublic:\n void Tester(int xx, int yy = 5)\n {\n x = xx;\n y = yy;\n cout << ++x % --y;\n }\n};\n\nint main()\n{\n CuriousTab objCuriousTab;\n objCuriousTab.Tester(5, 5);\n return 0;\n}\n

Introduction / Context:
The problem focuses on default parameters and operator semantics in C++. It asks you to evaluate an expression that uses pre-increment and pre-decrement with integer operands and the modulo operator, all inside a class method invoked with explicit arguments.

Given Data / Assumptions:

• Method Tester(int, int = 5) is called as Tester(5, 5).
• Computation: cout << ++x % --y; with integer x and y.
• Free function Tester declared above is unrelated to the member call.

Concept / Approach:
Pre-increment ++x yields 6. Pre-decrement --y yields 4. The modulo operation with integers computes 6 % 4, which equals 2. The method prints this result directly.

Step-by-Step Solution:

Verification / Alternative check:
Replacing pre- with post- (x++ % y--) would evaluate to 5 % 5 → 0 first, but the code explicitly uses pre-operators, so the printed result is 2.

Why Other Options Are Wrong:

• The program will print the output 0./1.: Do not match 6 % 4.
• The program will print the output garbage value.: All variables are well-defined.
• The program will report compile time error.: There is no ambiguity; member call resolves correctly.

Common Pitfalls:
Confusing the free Tester declaration with the class method; mixing up pre- and post- increments which change the numeric result; overlooking that modulo is only valid for integral types (which we have here).

Final Answer:
The program will print the output 2.