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 negative indexes, or indexes that don't begin with 0?
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.
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.
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