C dynamically growing array

I can use pointers, but I am a bit afraid of using them.

If you need a dynamic array, you can’t escape pointers. Why are you afraid though? They won’t bite (as long as you’re careful, that is). There’s no built-in dynamic array in C, you’ll just have to write one yourself. In C++, you can use the built-in std::vector class. C# and just about every other high-level language also have some similar class that manages dynamic arrays for you.

If you do plan to write your own, here’s something to get you started: most dynamic array implementations work by starting off with an array of some (small) default size, then whenever you run out of space when adding a new element, double the size of the array. As you can see in the example below, it’s not very difficult at all: (I’ve omitted safety checks for brevity)

typedef struct {
  int *array;
  size_t used;
  size_t size;
} Array;

void initArray(Array *a, size_t initialSize) {
  a->array = malloc(initialSize * sizeof(int));
  a->used = 0;
  a->size = initialSize;
}

void insertArray(Array *a, int element) {
  // a->used is the number of used entries, because a->array[a->used++] updates a->used only *after* the array has been accessed.
  // Therefore a->used can go up to a->size 
  if (a->used == a->size) {
    a->size *= 2;
    a->array = realloc(a->array, a->size * sizeof(int));
  }
  a->array[a->used++] = element;
}

void freeArray(Array *a) {
  free(a->array);
  a->array = NULL;
  a->used = a->size = 0;
}

Using it is just as simple:

Array a;
int i;

initArray(&a, 5);  // initially 5 elements
for (i = 0; i < 100; i++)
  insertArray(&a, i);  // automatically resizes as necessary
printf("%d\n", a.array[9]);  // print 10th element
printf("%d\n", a.used);  // print number of elements
freeArray(&a);