Is there a library function in c# for the mathematical modulus of a number - by this I specifically mean that a negative integer modulo a positive integer should yield a positive result.
edited to provide an example:
-5 modulo 3 should return 1
No this doesn't yield a positive result when doing -5 % 2
Snake
2010-04-22 13:12:59
He wants `-5 % 2` to be **+1**, not -1.
SLaks
2010-04-22 13:13:34
+8
A:
Try (a % b) * Math.Sign(a)
Try this; it works correctly.
static int MathMod(int a, int b) {
return (Math.Abs(a * b) + a) % b;
}
SLaks
2010-04-22 13:12:51
+1 for being the first to post a correct algorithm (but watch out for overflow!). To strictly answer his question, though, no, there's nothing built into the framework for this.
Craig Stuntz
2010-04-22 13:32:47
Accepted as the first correct algorithm, but I think I am more likely to use ((a % b) + b) % b
penguat
2010-04-22 13:41:30
Same issue as SLaks's answer; this is indeed positive, but not really a modulus.
Craig Stuntz
2010-04-22 13:21:05
Sorry, I had misunderstood the question. Your ((a % b) + b) % b seems the best solution.
Paolo Tedesco
2010-04-22 13:54:43
@penguat: Because if x is non-negative then there is no reason to go through the extra work.
Michael Petito
2010-04-22 13:41:57
I'm not sure which is more work on average, the conditional or the fixed addition and modulus, not that it matters in my case. This answer is correct, anyway. Thanks.
penguat
2010-04-22 13:58:09
@penguat: Hard to say; I always consider division and modulus operations to be relatively expensive (unless they are trivial, as in divide by 2 or mod 2). I imagine the overhead of the ternary operator is only a few cycles for the check and jumps. Plus, there is no chance for overflow with non-negative x in this computation. However, without the ternary operator, you could overflow if `x = 1` and `m = int.MaxValue` (or in any scenario where 0 < x < m and x + m > int.MaxValue).
Michael Petito
2010-04-22 15:18:58
I would definitely guess that the extra modulo is faster than the branch on modern processors. However, I've never tested it, so I don't really know. Also, it's a pretty slight distinction.I'm not sure which is more readable. I was going to say without the ternary operator at first, but when I think about it that x < 0 does tell the reader why we're doing something fancy without having the need for a comment.
clahey
2010-04-23 01:29:12
+3
A:
Well the definition (if I'm not mistaken) is something like this
a mod b = a - b * floor(a/b)
It's probably pretty slow and beware of integer division just like built in modulus :)
Other option is to modify the result of built-in modulus according to the signs of operands. Something like this:
if(a < 0 && b > 0)
{
return (a % b + b) % b;
}
else if ....
Maciej Hehl
2010-04-22 13:26:50
A:
Fix :
(ans=a%b)<0 ? (a<0 && b<0 ? (ans-b)%(-b) : (ans+b)%b) : ans
Mister_Egg_Head
2010-04-22 16:44:28
what about -10 mod 3? your solution would give -1 where the required answer is 2.
penguat
2010-04-22 17:47:37
@penguat a=-10 , b=3 ans= -1 < 0 , so solution = (-1+3) % 3 =2 what's wrong ?
Mister_Egg_Head
2010-04-22 17:57:17
A:
If you're using any of these algorithms and you need to do division also, don't forget to make sure that you subtract 1 when appropriate.
I.e.,
if -5 % 2 = -1 and -5 / 2 = -2, and if you care that -5 / 2 * 2 + -5 % 2 = -5, then when you calculate -5 % 2 = 1, that you also calculate -5 / 2 = -3.
clahey
2010-04-23 01:25:48
I'm not quite sure what you're getting at here... I'd normally preserve the original number rather than try to reconstruct it.
penguat
2010-04-23 10:17:47
I'm not suggesting you reconstruct it. I was just concerned that your algorithm might use the / operator as well and depend on them being consistent.
clahey
2010-04-23 15:26:53