In C, consider this program saved as sample.c: int main(int argc, char **argv) { argc = argc - (argc - 1); printf("%s", argv[argc - 1]); return 0; } Regardless of how many command-line arguments are supplied, what does this program print?

Introduction / Context:
This question focuses on command-line arguments in C and how argc and argv work. The program performs an unusual manipulation of argc before using it to index into argv. Understanding the algebraic simplification of the expression and the meaning of argv elements is key to predicting the output.

Given Data / Assumptions:

• argc is the count of command-line arguments, including the program name.
• argv is an array of pointers to strings, where argv[0] is the program name and argv[1]..argv[argc - 1] are additional arguments.
• The code reassigns argc with argc = argc - (argc - 1); and then prints argv[argc - 1].
• We assume argv is well-formed and argc is at least 1.

Concept / Approach:
We first simplify the assignment to argc and see what value it takes, regardless of its original value. Then we evaluate argv[argc - 1] with this new argc. Since argv[0] is defined as the program name, if argc becomes 1, argv[argc - 1] will always be argv[0], which is the program name string.

Step-by-Step Solution:
Step 1: Let the original value of argc be N (where N ≥ 1).Step 2: Compute the new argc by substituting: argc = argc - (argc - 1).Step 3: Algebraically, N - (N - 1) simplifies to N - N + 1 = 1.Step 4: Therefore, after the assignment, argc is always 1, no matter how many arguments were supplied initially.Step 5: The program then prints argv[argc - 1] which is argv[1 - 1] = argv[0], the program's own name.

Verification / Alternative check:
If you run the program as ./sample 10 20 30, the original argc is 4. After the assignment, argc becomes 1, and argv[0] still points to "./sample". If you run it with no extra arguments, argc starts as 1 and remains 1 after the assignment, so argv[0] is still the program name. In all cases, argv[0] is printed, confirming the analysis.

Why Other Options Are Wrong:
Option B claims that the last command-line argument is printed, which would be argv[argc - 1] using the original argc, not the modified value. Option C suggests the first argument after the program name, which would be argv[1], but the code uses argv[0]. Option D is wrong because argv[0] is a valid string containing the program path or name, so printing it does not yield an empty string.

Common Pitfalls:
Students sometimes forget that argc includes the program name and only think about user-supplied arguments. Another pitfall is not simplifying the assignment algebraically and trying to reason through specific examples only. Recognizing that argc is forced to 1 clarifies the behavior immediately.

Final Answer:
The program always prints the program's own name, that is, the string stored in argv[0].