Map a 2D array onto a 1D array

Question

I want to represent a 2D array with a 1D array. A function will pass the two indicies (x,y) and the value to store. These two indicies would represent a single element of a 1D array, and set it accordingly. I know the 1D array needs to have the size of arrayWidth × arrayHeight, but I don't know how to set each element.

For example, how do I distinguish (2,4,3) from (4,2,3)? I tried setting the array as the x*y, but 2*4 and 4*2 would result in the same spot in the array and I need them to be different.

Answer 1

You need to decide whether the array elements will be stored in row order or column order and then be consistent about it. http://en.wikipedia.org/wiki/Row-major_order

The C language uses row order for Multidimensional arrays

To simulate this with a single dimensional array, you multiply the row index by the width, and add the column index thus:

 int array[width * height];   int SetElement(int row, int col, int value)  {     array[width * row + col] = value;    } 

Answer 2

The typical formula for recalculation of 2D array indices into 1D array index is

index = indexX * arrayWidth + indexY; 

Alternatively you can use

index = indexY * arrayHeight + indexX; 

(assuming that arrayWidth is measured along X axis, and arrayHeight along Y axis)

Of course, one can come up with many different formulae that provide alternative unique mappings, but normally there's no need to.

In C/C++ languages built-in multidimensional arrays are stored in memory so that the last index changes the fastest, meaning that for an array declared as

int xy[10][10]; 

element xy[5][3] is immediately followed by xy[5][4] in memory. You might want to follow that convention as well, choosing one of the above two formulae depending on which index (X or Y) you consider to be the "last" of the two.