Thread: multmatrix()

Code:

m0 m4 m8 m12
m1 m5 m9 m13
m2 m6 m10 m14
m3 m7 m11 m15

Depending on the order of the matrix this function varies. However you should be able to get it.

Code:

for (int i=0;i<4;i++)
{
  for (int j=0;j<4;j++)
  {
    for (int k=0;k<4;k++)
    {
       matOut[i][j]+=mat1[i][k]*mat2[k][j];
    }
  }
}

Now unroll the inner loops.

matOut[i][j]=(mat1[i][0]*mat2[0][j])+(mat1[i][1]*mat2[1][j]).....

etc, etc.

You can also unroll the j loop with a bit more code.
Now once you unroll it, use the matrix I provided to eliminate all of the loops.

It's a strange advise considering matrixes are traditionally 2D containers. Also, from your own experience it becomes harder to code operations on matrixes built from a one dimensional array.

Regardless, if you insist on a 1D array, the solution is to know what are the bounds of each dimension. A 4x4 1D matrix is of size 16, with the second dimension starting at element 8 and ending at element 15

EDIT: Oops! My bad... Don't know where I had my head...

... with the second dimension starting at element 4 and ending at element 7, third starting at 8 and ending at 11, and so forth. Basically the offset of every dimension starting point is 4 indexes

Code:

size_t location[] =
{
    0, 4, 8, 12,
    1, 5, 9, 13,
    2, 6, 10, 14,
    3, 7, 11, 15,
};

opengl[ location[ x ] ] = whatever;

Lookup table.


Quzah.

Originally Posted by Bubba

That is a 1D array. Check my diagram.

Yup.

Originally Posted by linuxdude

but my array is only a one dimensional array. Every book that I have read has said to keep it this way as to not confuse with c's implementation of 2 dimensional arrays. Should I declare my array otherwise?

Quzah.