Pointer to argument string and prefix increment: what character is printed? /* myprog.c / #include <stdio.h> int main(int argc, char argv[]) { / Invoked as: myprog friday tuesday sunday / printf("%c", ++argv[1]); return 0; }

Introduction / Context: This program tests understanding of operator precedence and pointer arithmetic on C strings in argv. Specifically, it uses prefix ++ on a char (argv[1]) and then dereferences to print a single character.

Given Data / Assumptions:

• argv[1] points to the string literal "friday".
• The expression is ++argv[1].
• Operator precedence: [] binds first, so ++ applies to the result of argv[1] (a char), not to argv itself.

Concept / Approach: argv[1] is a char. Applying prefix ++ to a char moves it to the next character in the same string. After ++, the pointer points to the second character. Dereferencing with * yields that character.

Step-by-Step Solution:

Verification / Alternative check: Rewriting as char *p = argv[1]; ++p; putchar(*p); confirms the same result.

Why Other Options Are Wrong:

• f: That would be *argv[1] without increment.
• m or y: Characters from other strings or positions, not referenced by this expression.

Common Pitfalls: Misreading ++argv[1] as ++argv then [1]; forgetting [] binds tighter than ++; confusing pre-increment with post-increment.

Final Answer: r