Implement 2d array coordinates in 1d array in C++ - c++

Question

The code inside the for loop is for the x and y (j and i) "coordinates" from a 2d array. How could I implement this neighbor/index finding in a 1d array?
I think I could implement it for the first four equations. But i'm confused as how to implement up-left etc.
for(int i=0; i<cols*rows; i++){
//Counts current index's 8 neigbour int values
int count=0;
int x = i%cols;
int y = i/rows;
//rows y i
//cols x j
count+= [grid][i][(j-1+cols)%cols] //left
+[grid][i][(j+1+cols)%cols] //right
+[grid][(i-1+rows)%rows][j] //up
+[grid][(i+1+rows)%rows][j] //down
+[grid][(i-1+rows)%rows][ (j-1+cols)%cols] //up-left
+[grid][(i+1+rows)%rows][ (j+1+cols)%cols] //down-right
+[grid][(i+1+rows)%rows][ (j-1+cols)%cols] //down-left
+[grid][(i-1+rows)%rows][ (j+1+cols)%cols] ;//up-right
}

Answer 1

Starting with a 1-D vector:
int rows = 10;
int cols = 10;
vector<int> grid(rows * cols);
You can manage this in different ways, example
for(int y = 0; y < rows; y++)
{
for(int x = 0; x < cols; x++)
{
int point = grid[y * rows + x];
}
}
Where you can access any point at any given x and y in a 2-dimensional plane.
Top-left is:
x = 0;
y = 0;
bottom-right is
x = cols - 1;
y = rows - 1;
And so on.

Answer 2

Use a function like this
inline int idx(const int i, const int j, const int rows) const
{
return i * rows + j;
}
to convert the 2d indices to 1d indices.
This way you don't have to change your algorithm.
Usage would be grid[idx(i, (j-1+cols)%cols, rows)].

Answer 3

The basic formula for computing the 1d coordinate from the 2d index pattern is usually one of the following:
row_index * row_length + column_index
column_index * column_length + row_index
Which one applies to your case depends on whether you would like to have a row-based or column-based memory layout for your 2d array. It makes sense to factor out the computation of this index into a separate function, as suggested in the other answer.
Then you just need to fill in the values somehow.
You could do it like this, for example:
// iterate big picture
// TODO: make sure to handle the edge cases appropriately
for (int i_row = 1; i_row < n_rows - 1; i_row++) {
for (int i_col = 1; i_col < n_cols -1; i_col++) {
// compute values
dst[i_row*n_cols+i_col] = 0;
for (int r = i_row-1; r < i_row+2; r++) {
for (int c = i_col-1; c < i_col+2; c++) {
dst[i_row*n_cols+i_col] += src[r*n_cols + c];
}
}
}
}
Assuming src and dst are distinct 1d vectors of size n_rows*n_cols...