note 33903 deleted from language.operators.arithmetic by sniper

Note Submitter: me@somewhere.com ---- Since Modulo is defined, there is a way to define integer division such that it STAYS in integer mode and you never have to worry about float rounding problems. $int_quotient = (int) ( ( $a - $a % $b ) / $b ) The int is merely there for casting the result to a integer. Take 14 / 4 $a - $a % $b is integer math, and removes the remainder, this ensures a clean division... 14 - 14 % 4 = 14 - 2 = 12 <- Stays integer By the above step, removing the remainder guarantees the resulting division will yield a whole number, yes it is a float, but being a whole number, there is no funky floating point artifacts. All the math till this point has been integer. 12 / 3 = 4.0, which is converted to INT by the cast. Now, we get.... 4 and 2 as the integer results for 14/3 & 14%3. Is this correct? 4*3 + 2 = 14, so it works. Negative cases are funky, but I'm sure this can be modded to deal with 14/-3, and -14/3. 3