C++ (recursion with default arguments in a member function) — does the call terminate or loop forever, and what would be printed?\n\n#include<iostream.h>\nclass CuriousTab {\npublic:\n void CuriousTab(int x = 15) {\n x = x / 2;\n if (x > 0)\n CuriousTab();\n else\n cout << x % 2;\n }\n};\nint main() {\n CuriousTab objIB;\n objIB.CuriousTab();\n return 0;\n}\n\nChoose the correct behavior.

Introduction / Context:
This snippet highlights a common trap: recursion with a default parameter that is re-applied at each call. Although the body halves a variable and seems to move toward a base case, the recursive call does not pass that new value along, so the base case is never reached.

Given Data / Assumptions:

• Member function signature: void CuriousTab(int x = 15).
• Body halves x and recurses with CuriousTab() (no argument).
• Default argument value 15 is used whenever the call omits the parameter.

Concept / Approach:
Each recursive call with no explicit argument uses the default, setting x to 15 again, then halving to 7, then recursing again with the default, and so on. The halved value is never propagated into the next call, so the condition x > 0 remains true on every activation until the call stack overflows or the runtime aborts, i.e., it behaves like an infinite loop (non-terminating recursion).

Step-by-Step Solution:

Verification / Alternative check:
Fix by passing the updated value: CuriousTab(x). Then the sequence 15 → 7 → 3 → 1 → 0 will terminate, printing 0 % 2 = 0.

Why Other Options Are Wrong:

• Printing 1, 2, or 15 presumes termination; it does not terminate.
• Compilation error does not occur; the code is syntactically valid.

Common Pitfalls:
Confusing default arguments with persistent state, and forgetting that defaults are reapplied at each call when the argument is omitted.

Final Answer:
The program will go into an infinite loop.