Integer Arithmetics

[JonasKunze]

Currently there are some lines with similar code to the following in StrawL1Algorithm.cpp:

int j=(floor) (i/2);
(int) floor((i%8)/2.0);

The usage of floor in these cases does not change the result but only takes CPU time. To understand this you need to know some basics in integer arithmetics. In a few words this is:

• If a, b are integers, a/b is the integer with the largest absolute value which is smaller or equal to (a/(float)b): 3/2==1, -3/2==-1
• If f is a double or float, floor(f) is the largest integer (I'm not talking about the c type int) smaller or equal to f, stored in a double or float: floor(1.5)==1, floor(-1.5)==-2

So this means that i/2 is already an integer equal to (int)floor(i/2) because floor(j)==j for every j of type int.
Additionally i/2==(int)floor(i/2.0) if i>=0 (please note the 2.0 is now a double so i/2.0 is also a double). So using floor here is useless and only consumes about a factor of 10 times more CPU time:

int j = i/2; // faster version of int j=(floor) (i/2);
(i%8)/2; // faster version of (int) floor((i%8)/2.0);