C (Turbo C) — predict the output when using the Turbo C globals _argc and _argv. Command used: cmd> sample Jan Feb Mar /* sample.c */ #include<stdio.h> #include<dos.h> int main(int arc, char *arv[]) { int i; for (i = 1; i < _argc; i++) printf("%s ", _argv[i]); return 0; } What is printed?

Introduction / Context:This question tests knowledge of Turbo C's non-standard globals _argc and _argv and how they mirror the standard argc/argv. It also evaluates understanding of how command-line arguments are iterated and printed.

Given Data / Assumptions:

• Invocation: sample Jan Feb Mar.
• Loop prints _argv[i] for i = 1 to _argc - 1.
• Each token is followed by a trailing space in the printf format.

Concept / Approach:_argv[0] refers to the program name ("sample"). The loop begins at i = 1, so it prints only user-supplied arguments. The three arguments are "Jan", "Feb", and "Mar". The output will be those three separated by spaces (with a trailing space, which is visually harmless).

Step-by-Step Solution:

Verification / Alternative check:Changing the command to sample A B prints "A B". Starting at i = 0 would also print the program name, which this code intentionally avoids.

Why Other Options Are Wrong:

• No output: The loop iterates three times.
• sample Jan Feb Mar: The code starts from i = 1, not 0.
• Error: The code compiles in Turbo C where _argc/_argv are available.

Common Pitfalls:Assuming standard C only; here the code relies on Turbo C extensions but still runs as intended in that environment.

Final Answer:Jan Feb Mar