Print every argv element including program name: what is the full output order?
/* myprog.c */
#include
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;
}
-
A10 20 30
-
Bmyprog 10 20
-
Cmyprog 10 20 30
-
D10 20
Answer
Correct Answer: myprog 10 20 30
Explanation
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:
i = 0 → prints argv[0] = "myprog".i = 1 → prints "10".i = 2 → prints "20".i = 3 → prints "30".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