How can I use free when I used ** ( double pointer )? - c

Question

STUDENT ** list=NULL;
char *getid;
getid =(char*)malloc(sizeof(char) * 8);
printf("How many student? ");
int menuNum,num=0;
scanf("%d",&num);
list=(STUDENT**)malloc(num*sizeof(STUDENT*));
I used pointer like this.
As I learned from the professor, before finishing my code, I ought to use free() function to retrieve the allocated memory again.
Here s what i wanna ask you.
I learned that If i wanna use free() about (char *getid)
I know I should write
free(getid);
then How can I use free about
STUDENT ** list = NULL ; // **It's about struct**
Should I use like
free(list);
or
free(*list);
I think the former is right, but when I write like the latter, there's no error on my X-code.
Could you tell me about it ?

Answer 1

You should use:
free(list);
but when I write like free(*list);, there's no error on my X-code
That's why *list does not deallocate anything, since nothing is allocated there. However, your program suffers from a memory leak that way, since the memory pointed to by the double pointer named list is not free'd. However, you are just (un)lucky not to see your program crash.
If you used Valgrind though, you would see the memory leak.
Should I write malloc and write free() immediately? not in the end of the line?
No. You first allocate the memory dynamically with malloc(), then you use this memory (initialize, access, etc.) and when you do not need the memory anymore, and only then, you deallocate it with free().
However, if you had dynamically allocated space for list[i], then you should do:
free(list[i]);
free(list);
where the order matters, since you do not want to have tangling pointers!
Maybe my example on dynamically allocated a 2D array will more explanatory on this, even though it's a different data structure than a list.
PS: We don't cast what malloc returns.

Answer 2

You should free any memory allocated by malloc with free...
as you figured out, you requested 8 bytes of memory and saved the Pointer to it to your symbol getid (getid =(char*)malloc(sizeof(char) * 8);)
as for your list, this is a bit trickier:
You actually allocating a list of pointers that should point to other memory locations (those might be dynamically allocated as well)
list=(STUDENT**)malloc(num*sizeof(STUDENT*)); size of num pointers
Allocates the space for the list and saves the pointer to it at symbol list.
I'd remind that the memory allocated doesn't have to come initialized from the OS, so we should initialize it.
for(int i=0; i<num; i++)
{
list[i] = NULL;
}
You could also use memset(list, NULL, num * sizeof(STUDENT*)).
and you are very much correct you should free its memory by free(list),
But you should free the items in the list BEFORE you free the list of pointers itself. (again, if the items were dynamically allocated!)
for(int i=0; i<num; i++)
{
if(list[i] != NULL) // only if allocated.
{
free(list[i]); // free the element # position i
list[i] = NULL;
}
}