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

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