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?
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.
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()the allocated buffer.