Printing argv[0] only: observe program name output regardless of the numeric arguments. /* sample.c */ #include <stdio.h> int main(int argc, char argv[]) { / Invocations: sample 1 2 3 | sample 2 2 3 | sample 3 2 3 */ printf("%s ", argv[0]); return 0; }

Introduction / Context: argv[0] conventionally contains the program name used to invoke the executable. Printing argv[0] therefore shows the program name regardless of any extra arguments supplied by the user.

Given Data / Assumptions:

• Three different invocations are shown with various numeric arguments.
• The code always prints argv[0] and ignores others.
• Program binary is assumed to be named "sample" in the environment.

Concept / Approach: Since only argv[0] is printed, the output remains constant. Arguments beyond argv[0] do not influence the string being printed.

Step-by-Step Solution:

Verification / Alternative check: Replace %s with printing *argv to observe identical behavior, since *argv equals argv[0] initially.

Why Other Options Are Wrong:

• sample 3 2 3 / sample 1 2 3: These would require printing multiple arguments.
• Error: The code is valid and compiles.

Common Pitfalls: Expecting the entire command line to be printed by %s with argv[0]; confusing argv printing with environment-specific shells that echo commands.

Final Answer: sample