Dereference of incremented argument pointer: identify the printed character from argv[2]. /* sample.c / #include <stdio.h> int main(int argc, char argv[]) { / Invoked as: sample friday tuesday sunday / printf("%c", ++argv[2]); return 0; }

Introduction / Context: The expression ++argv[2] combines array indexing, prefix increment, and dereferencing on a pointer to char. Understanding operator precedence and the effect of incrementing a char is crucial.

Given Data / Assumptions:

• argv[2] points to the string "tuesday".
• ++argv[2] advances the pointer to the next character in the same string.
• then dereferences that advanced pointer to yield a single character.

Concept / Approach: [] binds more tightly than ++, so we first get argv[2] (a char), then apply prefix ++ to move within the string, and finally use * to read the character now pointed to.

Step-by-Step Solution:

Verification / Alternative check: Assign char *p = argv[2]; ++p; putchar(*p); to observe the same result without compact operators.

Why Other Options Are Wrong:

• s / r / f: Other characters from different positions or different strings; the code specifically targets the second character of argv[2].

Common Pitfalls: Misinterpreting ++argv[2] as ++argv then [2]; forgetting pre-increment advances before dereference.

Final Answer: u