note 34866 deleted from language.operators.arithmetic by sniper

Note Submitter: number_co@yahoo.com ---- saforres@alumni.SPAMGUARD.uwaterloo.ca told us that: "This sort of strategy will not work. Suppose $foo=9999999 and $bar=10000000. Then since $foo<$bar, you would want int($foo/$bar) to return 0, but 9999999/10000000+0.0000001=1.000000000, so (int)($foo / $bar + 0.0000001)=1." I have an idea that will get integer division to work. If you just do it like this: (int)(($foo+0.01)/$bar) (providing of course $foo is an integer as well), it'll work.