Re: So You Think You Can Const?

On 1/10/25 11:57 PM, Chris M. Thomasson wrote:

On 1/9/2025 11:40 PM, Keith Thompson wrote:

Andrey Tarasevich <andreytarasevich@hotmail.com> writes:

On 01/09/25 12:12 AM, Julio Di Egidio wrote:

I do not understand that: `free` is changing the pointed data, so
how can `const void *` even be "correct"?

>
`free` is destroying the pointed data.

>
Right.  In other words, it causes the pointed-to data to reach the end
of its lifetime.  "Changing" the data generally means modifying its
value (that's what "const" forbids).
>
Given:
>
     int *ptr = malloc(sizeof *ptr);
     *ptr = 42;
     printf("*ptr = %d\n", *ptr);
     free(ptr);
>
After the call to free(), the int object logically no longer exists.
Also, the value of the pointer object ptr becomes indeterminate.
Attempting to refer to the value of either ptr or *ptr has undefined
behavior.

 I must be missing something here. Humm... I thought is was okay to do something like this:
_____________________________
#include <stdio.h>
#include <stdlib.h>
 int main() {
     int* a = malloc(sizeof(*a));
      if (a)
     {
         *a = 42;
          printf("a = %p\n", (void*)a);
         printf("*a = %d\n", *a);
          free(a);
          printf("a = %p was just freed! do not deref\n", (void*)a);
     }
      return 0;
}
_____________________________
 Is that okay?
 [...]

No, because the value of a has become indeterminate, and operating on it, even to just look at its value, can trap.
you could save a representation of it either in a char array or as a uintptr_t value, and work with that (but not try to recreate a pointer with it, as that pointer "value" has become indeterminate).
This issue CAN occur if the implementation is using segment_tag + offset pointers, and free invalidates the segment_tag of that the pointer used, and the implementation will perhaps validate the segment_tag when looking at the pointer value. (perhaps pointers are loaded into registers that automatically validate the segment_tag in them).