C++ modulo is not defined for floating-point operands: identify the outcome when using % with float values in a method.\n\n#include <iostream.h>\n\nvoid Tester(float xx, float yy = 5.0); // unrelated free function\n\nclass CuriousTab\n{\n float x;\n float y;\npublic:\n void Tester(float xx, float yy = 5.0)\n {\n x = xx;\n y = yy;\n cout << ++x % --y; // invalid: % requires integral operands\n }\n};\n\nint main()\n{\n CuriousTab objCuriousTab;\n objCuriousTab.Tester(5.0, 5.0);\n return 0;\n}\n

Introduction / Context:
In C++, the % (remainder) operator is defined only for integral types. Attempting to apply it to floating-point operands is ill-formed. This snippet intentionally uses % with float values to provoke a compiler diagnostic.

Given Data / Assumptions:

• Method Tester takes float parameters and assigns to float members.
• Expression: ++x % --y with x and y of type float.
• Classic headers (iostream.h) do not affect the core language rule about %.

Concept / Approach:
Both operands to % must be integral. Since ++x and --y are still floats, there is no valid operator% overload, and the code fails to compile with an error similar to “invalid operands to binary %”.

Step-by-Step Solution:

Verification / Alternative check:
Changing the members to int (and passing integer literals) would allow % and yield a value. Alternatively, use fmod from for floating-point remainders: fmod(++x, --y).

Why Other Options Are Wrong:

• Any printed numeric output assumes successful compilation and execution, which does not occur.
• “Garbage value” implies runtime behavior; the failure is compile-time.

Common Pitfalls:
Assuming % behaves like mathematical modulus for floats, or forgetting to use fmod for double/float types.

Final Answer:
The program will report compile time error.