Array -> 2D array

How can I convert an array of 25 elements into a 5x5 2D array?

Create 2D array, then loop through start to end and set individuals elements.

You could cast the addresses:

Code:

---

typedef int array2d [5][5];

*(array2d *)&myArray1D;

---

Or without typedefs:

Code:

---

*(int (*)[5][5])&myArray1D

---

Or if passed to a function, simply cast the 1D array to void pointer.

But if you need a 2d array, why do you have a 1D array? You should probably be loading your data directly into a 2d array in the first place.

Quote:

---

But if you need a 2d array, why do you have a 1D array? You should probably be loading your data directly into a 2d array in the first place.

---

You never 'need' a 2D array per se.

Quote:

---

Originally Posted by Bubba

You never 'need' a 2D array per se.

---

What bubba means is that you can make any dimension array using math: The index of an X-dimension array is:

Code:

---

index = p0 + p1 * n0 ... pX *n(X-1) * n(X-2) ... n0;

---

where p{0..X} is the position of that dimension, and n{0..X} is the size of that dimension.

Example:

Code:

---

// Treat a 1d 1000 element array as a 3D [10][10][10] array.
int array[1000]; 
int x, y, z;
for(x = 0; x < 10; x++)
{
  for(y = 0; y < 10; y++)
  {
      for(z = 0; z < 10; z++)
      {
          array[x + (y * 10) + (z * 100)] = x * y * z;
      }
  }
}

---

This will be exactly what the compiler does internally anyways, when it deals with multidimensional arrays.

--
Mats