Fill a 2d array of pointers with pointers to structs in C - arrays

Question

I have a 2 dimensional array of pointers:
typedef struct Cell{
Position p;
unsigned int value;
} Cell;
typedef struct Position{
unsigned int x;
unsigned int y;
} Position;
int size = 4;
Cell ***cells = malloc(sizeof(Cell**) * size);
int i,j;
for(i = 0; i < size; i++){
cells[i] = malloc(sizeof(Cell*) * size);
for(j = 0; j < size; j++){
cells[i][j] = malloc(sizeof(Cell));
}
}
What I want to do now is fill this array with pointers to cells, and initialize these cells to contain the value 0 like this:
for(i = 0; i < size; i++){
for(j = 0; j < size; j++){
Position p = {i,j};
Cell c = {p, 0};
cells[i][j] = &c; //This doesn't work
}
}
As you can already tell, writing the address of c into the pointer cells[i][j] is less than ideal, since every pointer now points to the same address. However I don't know how to fill this array with pointers pointing to individual addresses.
I tried something like this:
cells[i][j]->value = 0;
which of course also doesn't work. Can anyone give me a hint on how I can solve my problem?

Answer 1

I'm not sure why you say cells[i][j]->value = 0; doesn't work, I believe it should.
This should work to set all the members.
Cell ***cells = malloc(sizeof(Cell**) * size);
for(i = 0; i < size; i++){
cells[i] = malloc(sizeof(Cell*) * size);
for(j = 0; j < size; j++){
cells[i][j] = malloc(sizeof(Cell));
cells[i][j]->p.x = i;
cells[i][j]->p.y = j;
cells[i][j]->value = 0;
}
}

Answer 2

Additional:
for(i = 0; i < size; i++){
for(j = 0; j < size; j++){
Position p = {i,j};
Cell c = {p, 0};
cells[i][j] = &c; //This doesn't work
}
// to this line, 'c' & 'p' will be freed since it was allocated on stack not heap.
// 'c' & 'p' will be freed when program gets out of 2nd for loop, so either you can
// use memcpy to copy 'c' into cells[i][j] or use malloc to allocate 'c' and 'p'
}

Answer 3

You've already gotten your answer but I see no obvious reason to do that many allocations unless sizeof(Cell) * size * size is large enough to make a single allocation fail (but allocating smaller parts doesn't). I suggest doing one allocation only which gives a few benefits. In no particular order:
Faster allocation in most cases.
Faster access to the elements that will be tightly packed and therefore be more cache friendly.
Easier memory management - there is only one pointer to check and free.
Less total memory allocated than when allocating memory for a lot of pointers.
Example 1 - Make cells a pointer to the first of size number of elements of Cell[size].
// one allocation:
Cell (*cells)[size] = malloc( size * sizeof *cells );
if(cells == NULL) exit(1); // check for allocation failure
for(int i = 0; i < size; i++){
for(int j = 0; j < size; j++){
cells[i][j].p.y = i;
cells[i][j].p.x = j;
cells[i][j].value = 0;
}
}
Example 2 - Make cells a pointer to a single Cell[size][size]:
// one allocation:
Cell (*cells)[size][size] = malloc( sizeof *cells );
if(cells == NULL) exit(1); // check for allocation failure
for(int i = 0; i < size; i++){
for(int j = 0; j < size; j++){
(*cells)[i][j].p.y = i;
(*cells)[i][j].p.x = j;
(*cells)[i][j].value = 0;
}
}