In C, argv points to the command-line argument vector. For the program below executed as: cmd> myprog one two three, what does the statement printf("%s\n", *++argv); print? /* myprog.c */ #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { printf("%s\n", *++argv); return 0; }

Introduction / Context:
This question probes pointer arithmetic with argv. The expression ++argv advances the pointer to the first user argument and then dereferences it to a char string.

Given Data / Assumptions:

• Command line: myprog one two three
• argv initially points at argv[0] (i.e., "myprog").

Concept / Approach:
Pre-increment ++argv moves the argv pointer from &argv[0] to &argv[1]. Dereferencing with * yields argv[1], the first user argument string, "one".

Step-by-Step Solution:

Verification / Alternative check:
If you used printf("%s\n", argv[2]); you would get "two". The *++argv form is a compact idiom for skipping argv[0].

Why Other Options Are Wrong:

• myprog: That would be *argv before increment.
• two or three: Those are later elements, not the first after increment.
• Undefined behavior: This is well-defined with argc > 1.

Common Pitfalls:
Confusing *argv++ with *++argv: post-increment returns the old pointer, pre-increment returns the incremented one; here pre-increment is used.

Final Answer:
one