Discussion
What is the output of this C code?
#include <stdio.h>
int x = 0;
void main()
{
int *const ptr = &x;
printf("%p\n", ptr);
ptr++;
printf("%p\n ", ptr);
}
Answer: B
Correct answer: Compile time error
int *const ptr declares a constant pointer to a modifiable int, so ptr itself may not be reassigned and ptr++ is rejected.
Contrast const int *ptr, where the const protects the pointed-to value and the pointer may still be moved. The position of the qualifier relative to the * decides which is fixed, and reading from the identifier outwards makes it clear: ptr is a const pointer to int.
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