dynamic array and pointer in C - c

Question

I'm a beginner in C and programming. I would like to ask some questions on dynamic array and pointer in C.
I am trying to create a dynamic array and increase its capacity, but I can't get my code working. I believe something is wrong in my setCapacityDynArr function.
Can someone give me some help?
Thanks!
struct DynArr {
TYPE *data; /* pointer to the data array */
int size; /* Number of elements in the array */
int capacity; /* capacity ofthe array */
};
void initDynArr(struct DynArr *v, int capacity) {
v->data = malloc(sizeof(TYPE) * capacity);
assert(v->data != 0);
v->size = 0;
v->capacity = capacity;
}
void freeDynArr(struct DynArr *v) {
if (v->data != 0) {
free(v->data); /* free the space on the heap */
v->data = 0; /* make it point to null */
}
v->size = 0;
v->capacity = 0;
}
int sizeDynArr(struct DynArr *v) {
return v->size;
}
void addDynArr(struct DynArr *v, TYPE val) {
/* Check to see if a resize is necessary */
if (v->size >= v->capacity) {
_setCapacityDynArr(v, 2 * v->capacity);
}
v->data[v->size] = val;
v->size++;
}
void _setCapacityDynArr(struct DynArr *v, int newCap) {
//create a new array
struct DynArr *new_v;
assert(newCap > 0);
new_v = malloc(newCap * sizeof(struct DynArr));
assert(new_v != 0);
initDynArr(new_v, newCap);
//copy old values into the new array
for (int i = 0; i < new_v->capacity; i++) {
new_v->data[i] = v->data[i];
}
//free the old memory
freeDynArr(v);
//pointer is changed to reference the new array
v = new_v;
}
int main(int argc, const char * argv[]) {
//Initialize an array
struct DynArr myArray;
initDynArr(&myArray, 5);
printf("size = 0, return: %d\n", myArray.size);
printf("capacity = 5, return: %d\n", myArray.capacity);
//Add value to the array
addDynArr(&myArray, 10);
addDynArr(&myArray, 11);
addDynArr(&myArray, 12);
addDynArr(&myArray, 13);
addDynArr(&myArray, 14);
addDynArr(&myArray, 15);
for (int i = 0; i < myArray.size; i++) {
printf("myArray value - return: %d\n", myArray.data[i]);
}
return 0;
}

Answer 1

//pointer is changed to reference the new array
v = new_v;
This is your problem, a classic mistake in C. In fact the function changes its own copy of the pointer, the caller never sees the change. The problem is amply described by this C FAQ.
I suggest a different approach. There's no reason to make a new v: you simply want more storage associated with it. So instead of actually changing v, you'll probably want to just call realloc on the storage: v->DATA.
You might get away with something like:
tmp = realloc(v->data, newCap * sizeof *v->data);
if (!tmp)
error;
v->data = tmp;
And this way you don't need to copy the elements either: realloc takes care of that.

Answer 2

//pointer is changed to reference the new array
v = new_v;
Your original pointer outside the function is not changed, since you passed the value of the pointer not the address of it here:
void _setCapacityDynArr(struct DynArr *v, int newCap)
{

Answer 3

Yes it's an error in _setCapacityDynArr. It's an error because you declare an DynArr structure on the stack, then you try to free it and assign a new pointer to it. That will not work, as items allocated on the stack can't be freed.
What you want to do is to reallocate only the actual data, not the whole structure. For this you should use the realloc function.
There are other problems with the function as well, like you assigning to the pointer. This pointer is a local variable so when the function returns all changes to it will be lost.