Arbitrary length multi-dimensional arrays

Newbie question here. It's been a while since I've done C programming,
and I hit a wall last night.

Let's say I have a three dimensional array, like so:

Code:

---

int p[2][3][3] =
{{{0,0,0},
{1,1,1},
{0,1,0}},

{{0,1,0},
{1,1,0},
{0,1,0}}};

---

I also have a number of other three dimensional arrays, generated with
a code generating script. The size of each dimension varies with each
one. Some are [5][3][3], some are [2][5][5], etc...

How would I declare a variable that could hold any of these
3-dimensional arrays? I'm tripping over the pointer syntax.

I'd like to be able to do something like:
int ***val = p; //The variable p from the last example.

I'm pretty sure that ***val is the wrong way to go about it.

Thanks for your help!

To start:
http://c-faq.com/aryptr/ary2dfunc3.html

One way would be dynamic memory allocation. Let's say your program knows how big each dimension is and they're stored in the variables s_dim1, s_dim2, s_dim3. You could do something like this:

Code:

---

{
  int ***p;
  int s_dim1 = 4, s_dim2 = 5, s_dim3 = 2;
  int i, j;

  p = malloc(sizeof(*p) * s_dim1);

  for(i = 0;i < s_dim1;++i)
  {
    p[i] = malloc(sizeof(**p) * s_dim2);

    for(j = 0;j < s_dim2;++j)
      p[i][j] = malloc(sizeof(***p) * s_dim3);
  }
}

---

There isn't any way to just make an array in C without giving it a size, so the dynamic memory allocation is probably the best way to go.

Disclaimer: I didn't test the above code, but it should work.

Here is code that's tested for two dimensional array

int **array1 = (int **)malloc(nrows * sizeof(int *));
for(i = 0; i < nrows; i++)
array1[i] = (int *)malloc(ncolumns * sizeof(int));

access allocated 2D array like: array1[1][2] = 5;

itsme86's answer is much better.
It avoids casting the result of malloc, and avoids explicit reference to sizeof(type).

Though it is possible to allocate all 3 dimensions with just 3 malloc's.
Here is a recent example for 2 dimensions, just to give you the idea.
http://forums.devshed.com/showpost.p...22&postcount=5