note 24913 deleted from language.operators.arithmetic by sniper

Note Submitter: saforres@alumni.SPAMGUARD.uwaterloo.ca ---- niekomaatjes@hotmail.com wrote: "To avoid this, you can add a real tiny fraction to the total, like this: (int)($foo / $bar + 0.0000001)" 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. It's not generally a good idea to rely on truncated floats for integer divisions.