Print every argv element including program name: what is the full output order?\n\n/* myprog.c */\n#include <stdio.h>\n\nint main(int argc, char *argv)\n{\n / Invoked as: myprog 10 20 30 */\n int i;\n for (i = 0; i < argc; i++)\n printf("%s\n", argv[i]);\n return 0;\n}\n

Introduction / Context:
This common argv-processing loop prints each C-string in the argv array from index 0 up to index argc - 1. Because argv[0] is the program name, it appears first in the output, followed by all user-supplied arguments.

Given Data / Assumptions:

• Invocation: myprog 10 20 30.
• Loop: for (i = 0; i < argc; i++) printf("%s\n", argv[i]);
• Each element is printed on its own line; conceptual order matters more than line breaks for the MCQ.

Concept / Approach:
The loop indexes from 0, so the very first printed string is argv[0], not argv[1]. Thus the printed sequence begins with "myprog" before the numeric arguments.

Step-by-Step Solution:

Verification / Alternative check:
Changing the loop to start at i = 1 would skip the program name; as written, it includes it.

Why Other Options Are Wrong:

• 10 20 30: Omits argv[0].
• myprog 10 20: Omits the final argument.
• 10 20: Omits both the program name and one argument.

Common Pitfalls:
Assuming argv[0] is not part of argc/argv; forgetting that printing begins at index 0 in this loop.

Final Answer:
myprog 10 20 30