How to map a 2D array in a 1D array in C++?

Question

Lets say, I have a MxN arrays: int *b; and int **c; where

values in b are stored by columns (from c) and I need to put values from c to b
values in b are stored by rows (from c) and I need to put values from c to b

I know that basicly I would do it like that:

j = index / N;
i = index - (j * M);

in order to convert a 1D index into a 2D co-ordinate but have a problem how to implement those 2 cases, 1) and 2)?

Consider this answer: http://stackoverflow.com/a/13937325/942596 , It shows how to use a 1D array as a 2D array. — andre, Jan 29 '13 at 18:33
@ahenderson: yes but is it a row-major order or column-major order there? cause I need both — Brian Brown, Jan 29 '13 at 18:37
@ahenderson: ok, thanks, so one less :D Maybe you know how to make a column-major order? — Brian Brown, Jan 29 '13 at 18:45
Can't you always get column-major from row_major by using the transpose. so, swap (column, row) values. — andre, Jan 29 '13 at 18:52

score 2 · Accepted Answer · answered Jan 29 '13 at 18:37

2

Let W be the width of the 2D array, and H its height. Then assuming row-major layout, the 1D index 'ix' relates to the 2D-index [x,y] as such:

ix = y*w + x;
y = ix / w;  // implicit floor
x = ix % w;

e.g.:

const int W = 3, H=2;
int m[H][W] = {{1,2,3}, {4,5,6}};
int* oneD = &m[0][0];
assert(oneD[1*W + 2] == m[1][2]); // element 6, y=1, x=2

answered Jan 29 '13 at 18:37

emilk

great, thank you. How about column-major, is it like this: oneD[1*H+2] == m[1][2]; ? – Brian Brown – Brian Brown Jan 29 '13 at 19:56

1 Answers1