free pointer by other pointer points same memory

Question

Is the code down below possible?

int *a_p = malloc(sizeof(int));
int *b_p = a_p;
free(b_p);         //Free a_b by using b_p
a_p = b_p = NULL;

I'm very confused because two pointers point same memory...

If that code is impossible, could you teach me why, please?

A few links that provide basic discussions of pointers may help. [Difference between char *pp and (char*) p?](https://stackoverflow.com/a/60519053/3422102) and [Pointer to pointer of structs indexing out of bounds(?)...](https://stackoverflow.com/a/60639540/3422102) (ignore the titles, the answers discuss pointer basics) — David C. Rankin, Dec 06 '20 at 07:47
Some function for exemple with a void ** parameter can allocate memory you must free. — Florent, Dec 06 '20 at 08:32
Yes, it is possible, both pointers point to the same allocated buffer, so any of them (but not both) can be used to `free()` the allocated buffer. — Luis Colorado, Dec 07 '20 at 17:52

score 3 · Accepted Answer · answered Dec 06 '20 at 08:00

What you have is perfectly fine.

You create a_p, allocate a memory block, and assign the block's address to a_p.
After int *a_p = malloc(sizeof(int));:

int *a_p                  int @ 0x1234
+------------+           +------------+
| 0x1234     |           | ???        |
+------------+           +------------+

You create b_p with the same value as a_p.
After int *b_p = a_p;:

int *a_p                  int @ 0x1234
+------------+           +------------+
| 0x1234     |           | ???        |
+------------+           +------------+

int *b_p
+------------+
| 0x1234     |
+------------+

Notably, this didn't allocate any memory other than b_p itself.

You free the memory block you previously allocated.
After free(b_p);:

int *a_p
+------------+
| 0x1234     |
+------------+

int *b_p
+------------+
| 0x1234     |
+------------+

You overwrite the addresses.
After a_p = b_p = NULL;:

int *a_p
+------------+
| NULL       |
+------------+

int *b_p
+------------+
| NULL       |
+------------+

The two remaining blocks (a_p and b_p) have automatic storage duration, so they will be automatically freed when the function returns.

score 1 · Answer 2 · answered Dec 06 '20 at 07:42

Pointers are values. They can be copied via assignment.

In this example, b_p is set to the value of a_p and then free(b_p); is called.

At that moment, the value in b_p is one in a_p, which is a pointer returned from malloc() and not yet freed. Therefore, this is valid.

Another possibility is that malloc() failed and NULL is returned, but free(NULL); is valid and defined to do nothing, so the code is valid also in this case.

free pointer by other pointer points same memory

2 Answers2