C after free — what does printing the pointer value show? #include<stdio.h> #include<stdlib.h> int main() { int *p; p = (int ) malloc(20); / Assume, for illustration, p held address 1314 / free(p); printf("%u", p); / Printing pointer with %u is itself non-portable / return 0; } What will be the output conceptually?

Introduction / Context:This question examines behavior after freeing dynamically allocated memory. It also highlights that printing pointers with %u is non-portable (the correct specifier is %p). The central concept is undefined behavior and dangling pointers.

Given Data / Assumptions:

• p is freed using free(p).
• The code then prints p's value with %u.
• Assumed pre-free value 1314 is for illustration only.

Concept / Approach:After free, p becomes a dangling pointer: its stored bits may remain the same, but using it is undefined behavior. Even reading or printing it in a non-portable manner can produce unpredictable results. Therefore, you cannot rely on seeing 1314 or any specific value. The robust practice is to set p = NULL after free and print with %p if needed.

Step-by-Step Solution:

Verification / Alternative check:Setting p = NULL after free and then printing with printf("%p", (void)p) yields a consistent "(nil)" style representation on many systems, demonstrating defined behavior.

Why Other Options Are Wrong:

• 1314/1316: Suggests a deterministic address; not guaranteed.
• Random address: While colloquially similar, "Garbage value" better reflects UB in MCQ context.

Common Pitfalls:Believing that free changes the pointer value or that printing a freed pointer is safe. Always invalidate the pointer explicitly.

Final Answer:Garbage value