Thread: Array -> 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.

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