How do you free a 2D malloc'd array in C? - c

Question

I'm creating a 2D Array in C; am I freeing it correctly?
// create
int n = 3;
int (*X)[n] = malloc(sizeof(int[n][n]));
// set to 0
for(int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
X[i][j] = 0;
}
}
// is this all I need?
free(X);

Answer 1

You must call free once for each malloc. Your code has one malloc and one free (with the same address) so it is correct.

Answer 2

For every malloc(), there must be a matching free(). So if you call malloc() in a loop, there must be a similar loop later on that calls free() just as many times. Your code has one malloc() and one corresponding free(), so yes, you have freed everything you need to.
The most common way I've seen to allocate a two-dimensional array is to do something like:
int **arr = malloc(sizeof (int *) * num_rows);
for (i = 0; i < num_rows; ++i)
arr[i] = malloc(sizeof (int) * num_cols);
/*
* Using `sizeof *arr` and `sizeof **arr`
* respectively would have been more safe,
* but for the purpose of this example, I think
* using `int *` and `int` is simpler to
* understand.
*/
then later to free:
for (i = 0; i < num_rows; ++i)
free(arr[i]);
free(arr);
This makes the outer dimension hold num_row pointers to pointers to int (each of the num_row pointers points to the starting address of the 'column'). So each element in the outer dimension points to a row, and in each row there are num_cols elements in the inner dimension (the 'columns'), which is just a group of num_cols integers. Does this make sense to you? So you have to allocate num_rows integer pointers, and each one of these points to the first 'column' of that row--for each row you have to allocate space for num_cols integers, so you can see the loop will make a total of num_rows * num cols integers, and they way they are allocated allows you to use array indexing notation (two-dimensional in this case--each element in the outer dimension points to the start of a 'row', which contains a pointer to the start of a 'column', hence the double pointer) to access the elements. I know this is probably confusing, that is why I tried to describe it so many different times/ways, but please just ask any questions, especially about what you don't understand in particular, and I will be more than happy to work with you in helping understand it.
I'm not saying you have to create your 2-D array this way; your way works fine too, I just wanted to show you this way since you are likely to run into it since it's very common.
Also, look into Valgrind. You can use that tool to see if you have forgotten any free()s / have an unmatched malloc(), and lets you know if there is any allocated memory unreachable/leaked (maybe the pointer was changed before the call to free() so the call doesn't properly free the memory and you get a leak).

Answer 3

RastaJedi - another couple of things to note:
After a call to free(), it's a good idea to set the free'd variable to NULL, which helps prevent use after free.
Also, if you malloc in one block of code, then free in some called function, and then return from that function, sometimes C forgets about the final free (in your example, free(arr) ), and you should set the variable to NULL upon return.
I came here, and verified that the malloc and free calls that I wrote in the code I'm working on are indeed exactly as your "common" way, but kept running into a Seg Fault, because I was using malloc from in main, free-ing from a helper function which received the double-indirect head of the array. Having that all wrapped in a while-loop for user interaction, a second pass around was checking the array head for NULL-ness, but not seeing it as NULL, despite explicitly setting to NULL inside the helper function. Crawling thru the code with a debugger is how I identified the behavior, and why I came looking.
After being free'd, the memory is available, and variously (and randomly) reassigned; when the function with the free() calls returns, the memory that was allocated has been released, however it seems that the calling code still holds a pointer to that place in memory. There then exists a possible race-condition; by the time the calling code regains control, there's no guarantee as regards the location pointed to by the variable used for the array head, and therefore it's a very good idea to set the variable to NULL after the function returns.
Although you can use malloc'd pointers for pass-by-reference behavior, after a call to free() inside some function, the by-reference behavior breaks (because the thing being referenced is what has been free'd).
So, thank you for your answer, which confirmed that I was looking at the wrong part of my problem - just figured maybe someone else may stumble here, and possibly have the same issue I did.