In C, address-of and dereference cancellation with a char pointer: what is printed? #include<stdio.h> int main() { char p; p = "hello"; printf("%s ", &&p); return 0; }

Introduction / Context:This short puzzle shows that applying address-of (&) and dereference () operators in alternating fashion can cancel out, if they are applied in a type-consistent way. The format string %s expects a char* to a null-terminated string.

Given Data / Assumptions:

• p points to the string literal "hello".
• The expression is &&p (equivalent to p).
• We print with printf("%s", ...).

Concept / Approach:Reading right-to-left for unary operators: the innermost is &p (address of p), then (&p) which yields p, then &((&p)) which is &p again, and finally (&p) which returns p. So the entire expression is simply p, a char to the literal.

Step-by-Step Solution:Evaluate &&p → pprintf receives p as a char*It prints the full null-terminated string: hello

Verification / Alternative check:Replacing the expression with just p yields identical output.

Why Other Options Are Wrong:(a), (c), and (d) suggest starting at later characters; nothing in the code increments the pointer. (e) There is no invalid access here; reading a string literal via %s is fine (modification would not be).

Common Pitfalls:Assuming address-of and dereference operators always change the value; when paired correctly, they cancel out.

Final Answer:hello