C: howto convert a pointer to access it as a multidimensional array - c

Question

I do have a function call like:
int Filter(short* array, short nNumRow, short nNumCol)
but inside it I want to handle array like that:
array[y][x] = xx;
I try to solve this by declaring an array
short help[nNumRow][nNumCol];
help = array;
but this doesn't work that way. How can I handle that problem without changing function parameter list (this *array is result of a different function that I can't change)? Best of course would be not a copy (of memory) is needed.
Probably another option would be
array[y*nNumCol + x] = xx;
but I don't like this calculations. So how to do this best?
Thanks!

Answer 1

How can I handle that problem without changing function parameter list?
If you can't do that, then you are stuck with the "mangled array" array[y*nNumCol + x] notation (which is old style but otherwise ok).
The best and correct solution is to change the function to this:
int Filter (short nNumRow, short nNumCol, short array[nNumRow][nNumCol])
{
...
array[x][y] = something;
}
The last resort, which I would not recommend unless you are maintaining some old crap that can't be changed, is a dirty pointer conversion inside the function. Writing such code requires that you to know exactly what you are doing, because if the types of the actual data or the alignment mismatch, you will get very strange bugs. The below code works and is safe as far as the C language is concerned, but it isn't pretty:
// BAD CODE, avoid this solution
#include <stdio.h>
int Filter (short* array, short nNumRow, short nNumCol)
{
short(*array2D)[nNumCol]; // pointer to array of type short[nNumCol]
array2D = (short(*)[nNumCol])array; // dirty pointer conversion
for(int i=0; i<nNumRow; i++)
{
for(int j=0; j<nNumCol; j++)
{
printf("%d ", array2D[i][j]);
}
printf("\n");
}
return 0;
}
int main (void)
{
short array[2][2] = { {1,2}, {3,4} };
Filter((void*)array, 2, 2);
}

Answer 2

The best (optimal) way to do that is your own solution:
array[y*nNumCol + x] = xx;
Fo "beauty" reasons, you may use a function-like macro to access that data:
#define arrElement(array,x,y) ((array)[(y)*nNumCol + (x)]]))
If you need to apply this trick to only one array, then you can simplify the macro:
#define arrElement(x,y) (array[(y)*nNumCol + (x)]]))
If the size of the array is not known before the function call, you will need to add some complexity to the macro:
#define arrElement(x,y,nNumCol) (array[(y)*(nNumCol) + (x)]]))
Note: I did not text exactly the statements above, but I used the trick in the past several times, successfully.
You can use a pointer-type (as per your question) only if the array always has the same size. Otherwise, you will have to define the pointer-type dynamically at run-time, which is somewhere between difficult and impossible.
A sane thing to do is to pass to the function the array size also, and check if the coordinates actually fall inside the array. Otherwise, you may run into undefined behavior, accessing data outside defined range.

Answer 3

Not the most efficient way, but you can create a two dimensions array inside your function and copy the original array to it:
#include <stdio.h>
int Filter(short* array, short nNumRow, short nNumCol) {
short arr[nNumRow][nNumCol];
memcpy(arr, array, nNumRow * nNumCol * sizeof(short));
for (int i = 0; i < nNumRow; i++) {
printf("| ");
for (int j = 0; j < nNumCol; j++) {
printf("%d ", arr[i][j]);
}
printf("| \n");
}
return 0;
}
int main(void) {
short arr[] = {1, 2, 3, 4, 5, 6 ,7, 8};
Filter(arr, 2, 4);
printf("---\n");
Filter(arr, 4, 2);
return 0;
}
See it running here: https://ideone.com/58KhYj