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

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", 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