How allocate or free only parts of an array?

Question

See this example:

int *array = malloc (10 * sizeof(int))

Is there a way to free only the first 3 blocks?

Or to have an array with n

Answer 1

You can't directly free the first 3 blocks. You can do something similar by reallocating the array smaller:

/* Shift array entries to the left 3 spaces. Note the use of memmove
 * and not memcpy since the areas overlap.
 */
memmove(array, array + 3, 7);

/* Reallocate memory. realloc will "probably" just shrink the previously
 * allocated memory block, but it's allowed to allocate a new block of
 * memory and free the old one if it so desires.
 */
int *new_array = realloc(array, 7 * sizeof(int));

if (new_array == NULL) {
    perror("realloc");
    exit(1);
}

/* Now array has only 7 items. */
array = new_array;

As to the second part of your question, you can increment array so it points into the middle of your memory block. You could then use negative indices:

array += 3;
int first_int = array[-3];

/* When finished remember to decrement and free. */
free(array - 3);

The same idea works in the opposite direction as well. You can subtract from array to make the starting index greater than 0. But be careful: as @David Thornley points out, this is technically invalid according to the ISO C standard and may not work on all platforms.

Answer 2

You can't free part of an array - you can only free() a pointer that you got from malloc() and when you do that, you'll free all of the allocation you asked for.

As far as negative or non-zero-based indices, you can do whatever you want with the pointer when you get it back from malloc(). For example:

int *array = malloc(10 * sizeof(int));
array -= 2;

Makes an array that has valid indices 2-11. For negative indices:

int *array = malloc(10 * sizeof(int));
array += 10;

Now you can access this array like array[-1], array[-4], etc.

Be sure not to access memory outside your array. This sort of funny business is usually frowned upon in C programs and by C programmers.