Re: How to dynamically calculate an index?

[anonymouse via Digitalmars-d-learn]

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

Re: How to dynamically calculate an index?

[anonymouse via Digitalmars-d-learn]

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

How to dynamically calculate an index?

[ananymouse via Digitalmars-d-learn]

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