So I started playing around with C and having a lot of fun thus far.
There are however a couple of things I can not wrap my head around.
I know that this will end up on the stack
int i = 0;
I know this will reserve space for an integer on the heap and return the adress
int *i = malloc(sizeof(int));
However. If i do this
int i_one = 1, i_two = 2;
int *arr = calloc(2, sizeof(int));
arr[0] = i_one;
arr[1] = i_two;
i_one and two are stack allocated while arr is on the heap. Does this mean that arr will copy the values of i_one and two on to the heap or will it simply hold 2 references to variables on the stack. Im assuming its a variant of alt one considering (if I'm not misstaken) my stack allocated ints will be freed as soon as I exit this function.
So to summarize, when creating a dynamically allocated array using calloc. Do the entries in the array need to be pointers / heap allocated as well? In my head that would not make sense because then wouldnt i create an array of int pointers instead? And yes I know the size of a pointer is the same as an int so the example is a bit stupid but you get the point.
Thank you
The assignment operator is designed to assign a value stored in one object to another object.
So in these assignment statements
arr[0] = i_one;
arr[1] = i_two;
values stored in the variables i_one and i_two are copied into the memory occupied by the array elements arr[0] and arr[1]. Now if you will change for example the value stored in the variable i_one then the value stored in arr[0] will not be changed.
If you want to store references to objects i_one and i_two in the heap then you should write
int **arr = calloc(2, sizeof(int *));
arr[0] = &i_one;
arr[1] = &i_two;
Now you can change for example the value stored in i_one by means of the array element arr[0] the following way
*arr[0] = 10;