In C++ (old-style iostream.h), a function with a default argument is called once and computes a factorial-like product. Predict the printed result. #include <iostream.h> long FactFinder(long = 5); int main() { for (int i = 0; i <= 0; i++) cout << FactFinder() << endl; // calls FactFinder(5) return 0; } long FactFinder(long x) { if (x < 2) return 1; long fact = 1; for (long i = 1; i <= x - 1; i++) fact = fact * i; // multiplies up to x-1 return fact; }

Introduction / Context: This question checks your understanding of default function arguments, loop bounds, and how an off-by-one choice in the loop makes the routine compute a product up to x - 1 rather than x. It also reinforces that the loop in main executes exactly once and calls the function with its default parameter value.

Given Data / Assumptions:

• Function prototype: long FactFinder(long = 5).
• Main calls FactFinder() once due to for (i = 0; i <= 0; i++).
• In FactFinder, if x < 2 return 1; otherwise multiply i from 1 to x - 1.
• Classic headers (iostream.h) are assumed to compile in the intended environment.

Concept / Approach: Because the call uses the default, x is 5. The function multiplies numbers from 1 through 4 (not 5). That computes 1 * 2 * 3 * 4 = 24. It then returns this value, which is printed by cout from main, followed by a newline.

Step-by-Step Solution:

Verification / Alternative check: If the loop were i <= x instead of i <= x - 1, the result would be 120 for x = 5. The present code intentionally stops early.

Why Other Options Are Wrong:

• The program will print the output 1.: This would require x < 2 or an empty product, which is not the case.
• The program will print the output 120.: That would need multiplication up to 5.
• The program will print the output garbage value.: All variables are initialized; no undefined reads occur.
• The program will report compile time error.: The code is valid in the assumed toolchain.

Common Pitfalls: Confusing default-argument usage, assuming factorial(x) instead of factorial(x - 1), or thinking the for loop in main runs multiple times. Also, overlooking that the code prints once because the loop bound is i <= 0.

Final Answer: The program will print the output 24.