Double dereference on incremented argv pointer: which character prints first? /* myprog.c */ #include <stdio.h> int main(int argc, char *argv) { / Invoked as: myprog one two three */ printf("%c ", ++argv); return 0; }

Introduction / Context: The snippet highlights pointer-to-pointer manipulation with argv. It advances the argv pointer itself and then dereferences twice to reach the first character of the next argument string.

Given Data / Assumptions:

• argv initially points to argv[0] ("myprog").
• After ++argv, it points to argv[1] ("one").
• ++argv dereferences to the first char of argv[1].

Concept / Approach: ++argv increments the pointer-to-pointer so that argv becomes the next char, which is the C-string for the first argument. A second dereference (**) then gives the first character of that string.

Step-by-Step Solution:

Verification / Alternative check: Printing %s with *++argv would display "one"; switching to %c and ** prints its first character.

Why Other Options Are Wrong:

• myprog one two three / myprog one / two: These would require printing strings, not a single character.

Common Pitfalls: Confusing incrementing argv versus incrementing an element like argv[1]; mixing up * and ** dereference levels.

Final Answer: o