C - free() is not freeing the entire array

Question

I have a pointer to int, and after I call free, I see that only the first two elements get freed, the rest remain the same. Can anybody explain?

int main() {
        int* a = (int*)malloc(10*sizeof(int));

        a[0] = 12;
        a[1] = 15;
        a[2] = 100;
        a[3] = 101;
        a[4] = 102;
        a[5] = 103;
        a[6] = 109;
        a[7] = 999;

        printf("%d %d %d %d %d %d %d %d\n", a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7]);

        free(a);
        printf("Done freeing.\n");
        printf("%d %d %d %d %d %d %d %d\n", a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7]);

        return 0;
}

Output: (sorry for putting all the output in one line)

12 15 100 101 102 103 109 999 Done freeing. 0 0 100 101 102 103 109 999

Answer 1

When you invoke free, you're just telling the OS that you're done with that memory chunk. You promise not to use it again. If you break that promise, that's your problem.

Important sidenote:

Invoking if(ptr) or if(NULL != ptr) does NOT test if ptr points at an allocated memory chunk. It only tests whether it points at anything at all. Invoking free does NOT assign null to any pointers. You have to take care of that manually. I learned this lesson the hard way...

Important sidenote 2:

There's no general way to find out whether the memory ptr points at is allocated or not. Some platforms offer this feature, but not all.

Now, maybe you wonder, wouldn't it be better if free(ptr) assigned NULL to ptr? Maybe, maybe not. The downside would be that it's false safety. Consider this code:

int *p1=malloc(sizeof *p1);
int *p2=p1;
free(p1);

What value do you expect p2 to have? It's better to learn what it actually means that a pointer is/isn't NULL.