In C command-line arguments, the program below prints argv[argc - 1]. If it is executed as: cmd> sample (where argv[0] numerically refers to "sample"), what will it print? #include<stdio.h> int main(int argc, char **argv) { printf("%s ", argv[argc-1]); return 0; }

Introduction / Context: The item checks the layout of command-line arguments in C and how argc and argv relate. Specifically, argv[0] conventionally stores the program name (or path), and argc is the count of items in argv.

Given Data / Assumptions:

• Program prints argv[argc - 1].
• Program executed as: sample (no extra arguments).
• argv[0] corresponds to "sample" under normal conventions.

Concept / Approach: When the program is launched with no additional arguments, argc is 1 and argv points to an array of length 1 plus the terminating NULL. Therefore, argc - 1 equals 0, and the expression prints the string stored at argv[0], i.e., the program name path or name.

Step-by-Step Solution:

Verification / Alternative check: Add a second argument (e.g., sample X); now argc=2 and argv[argc - 1] = argv[1] = "X". The logic matches across cases.

Why Other Options Are Wrong:

• 0, samp: Not the value at argv[0].
• No output: The program prints a line; it does not exit silently.
• Undefined behavior: Index is valid when argc > 0.

Common Pitfalls: Forgetting that argv[0] is occupied by the program name; users sometimes assume argv[0] is the first user argument, which is incorrect.

Final Answer: sample