Re^3: Best way to make sure a number is an even multiple of another?

I don't trust modulo on a negative number. I'm never sure what value -2 % 5 will have. 3 or -2? I believe it depends on the language.

So, I want an easy, reliable way to add a multiple ($i) of $m so that $i - $n >= 0. The exact number doesn't matter, as

   ($r*$m + $s) % $m == $s % $m
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for any non-negative integers $r and $s, and positive integer $m.

$m*$n is both a multiple of $m, and at least as large as $n.

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Re^4: Best way to make sure a number is an even multiple of another? by Roy Johnson (Monsignor) on Oct 21, 2004 at 16:39 UTC
I think, then, that you want `$n + ($m - $n % $m) % $m;` [download] That ensures that what you are subtracting from $m is not larger than $m, and the result is the same, modulo $m. In Perl, modulo works as it is supposed to. `:-)` The result is between zero and the right operand. In Java, it's broken. This allows us to simplify the expression a bit further: `$n + -$n%$m;` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re^5: Best way to make sure a number is an even multiple of another? by hv (Prior) on Oct 22, 2004 at 10:44 UTC
The modulo operator works as I'd expect as long as you don't `use integer`; as soon as you do, you are at the mercy of your C implementation, which I believe is likely to vary between machines. On this Linux machine, for example: `zen% perl -le 'print -1 % 3' 2 zen% perl -Minteger -le 'print -1 % 3' -1 zen%` [download] If you're happy to rely on it, the original request to get the least multiple of $m >= $n can be satisfied by: `$n + ((-$n) % $m)`. Hugo	[reply] [d/l] [select]