• ananymouse (16/16) Jul 12 2022 Given a 3D matrix of shape[3, 5, 7] that is implemented using a
• anonymouse (23/27) Jul 12 2022 My current implementation:
• anonymouse (12/15) Jul 12 2022 Wait, I think I've got it.

ananymouse <anony mouse.com> writes:

Given a 3D matrix of shape[3, 5, 7] that is implemented using a 
1D array, I can calculate the index of any element (e.g. [1, 0, 
6]) as follows:

     index = 1 * 5 * 7 + 0 * 7 + 6

In general, if I have a K x L x M matrix, I can access 
index[i][j][k] by calculating i*L*M + j*M + k to yield the index. 
Likewise, when working with a 4D matrix[h][i][j][k], I can 
achieve the same by calculating h*L*M*N + i*M*N + j*N + k.

If I have both the index ([i, j, k]) and the shape ([K, L, M]) 
stored in two separate arrays of equal length, how could I 
dynamically generate the above calculations for any two such 
arrays of arbitrary length 3 or greater?

Been racking my brain on this for hours. Please point me in the 
right direction.

Thanks,
--anonymouse

anonymouse <anony mouse.com> writes:

On Tuesday, 12 July 2022 at 14:17:36 UTC, ananymouse wrote:
 Been racking my brain on this for hours. Please point me in the 
 right direction.

 Thanks,
 --anonymouse

My current implementation:

     ``` d
     // Allow for indexing to read a value, e.g a[0,0,0]
     T opIndex(A...)(A a) {
         static if(a.length == 3) {
             int index = a[0] * dims[1] * dims[2] + a[2];
             return data[index];
         }
         else static if(a.length == 2) {
            return data[a[0] * dims[1] + a[1];
         }
         else static if (a.length == 1) {
            return data[0]
         }
         assert(0);
     }
     ```

This does not scale well. Therefore, I'd like to change the first 
if statement to work with any array.length > 2 and dynamically 
generate the code. I've tried using a foreach loop to achieve 
this but failed miserably.

--anonymouse

anonymouse <anony mouse.com> writes:

On Tuesday, 12 July 2022 at 14:55:47 UTC, anonymouse wrote:
 I've tried using a foreach loop to achieve this but failed 
 miserably.

 --anonymouse

Wait, I think I've got it.

This little snippet does the trick:

     int index;
     foreach(i, v; a) {
         int t = v;
         foreach(d; dims[i+1 .. a.length])
             tmp *= d;
         index += tmp;
     }

Thanks, everyone.
--anonymouse