Reading and printing a C string via a pointer-to-const: With const char *s = "" then s = str where str is "Hello", what does the loop while(*s) printf("%c", *s++); print?

Introduction / Context:In C, a pointer declared as const char * may point to a character array, and you are allowed to move the pointer and read characters through it; the const qualifier forbids writing through the pointer, not reassigning the pointer itself. This question checks understanding of pointer iteration over a NUL-terminated string using a simple while loop that stops at the terminating '\0'.

Given Data / Assumptions:

• const char *s = ""; then char str[] = "Hello"; s = str;
• Loop: while (*s) printf("%c", *s++);
• Standard C string semantics: arrays are NUL-terminated.

Concept / Approach:

• The condition *s is true for any non-zero character; it becomes false at the NUL terminator.
• The expression *s++ prints the current character and then advances the pointer to the next character (post-increment of the pointer).
• No writes occur through s, so const-correctness is respected.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• Error/Undefined behavior: no invalid operations occur; only reads and pointer increments.
• 'H' or 'Hel': would require an early exit; the loop continues until NUL.

Common Pitfalls:

• Misreading *s++ as (*s)++ (which would attempt to modify the character); precedence rules mean it is *(s++), a pointer increment, not a character increment.
• Forgetting that the empty string "" initially is harmless since s is later reassigned to str before use.

Final Answer: