Perl
print 2 % -18;
-->
-16
Tcl
puts [expr {2 % -18}]
-->
-16
but VBScript
wscript.echo 2 mod -18
-->
2
Why the difference?
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bugmagnet
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Thanks for sharing this. I've tested in Python, Ruby, Javascript and bash now. JS and bash both said 2 and the other said -16. – PEZ Jan 16 '09 at 13:55
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Modulo operations mainly belong to Counting numbers and are arbitrarily defined outside of that. – Axeman Jan 16 '09 at 14:46
4 Answers
31
The wikipedia answer is fairly helpful here.
A short summary is that any integer can be defined as
a = qn + r
where all of these letters are integers, and
0 <= |r| < |n|.
Almost every programming language will require that (a/n) * n + (a%n) = a. So the definition of modulus will nearly always depend on the definition of integer division. There are two choices for integer division by negative numbers 2/-18 = 0 or 2/-18 = -1. Depending on which one is true for your language will usually change the % operator.
This is because 2 = (-1) * -18 + (-16) and 2 = 0 * -18 + 2.
For Perl the situation is complicated. The manual page says: "Note that when use integer is in scope, "%" gives you direct access to the modulus operator as implemented by your C compiler. This operator is not as well defined for negative operands, but it will execute faster. " So it can choose either option for Perl (like C) if use integer is in scope. If use integer is not in scope, the manual says " If $b is negative, then $a % $b is $a minus the smallest multiple of $b that is not less than $a (i.e. the result will be less than or equal to zero). "
Nick Fortescue
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Re "can do anything", no. If use integer is in scope it can only do what C does, and the C standard explicitly requires (a/n) + (a%n) = a but allows either choice for integer division. – ysth Jan 19 '09 at 11:26
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@B T, oops, missed out a factor of n, thanks for spot, I'll fix it – Nick Fortescue Apr 30 '10 at 15:40
9
Wikipedia's "Modulo operation" page explains it quite well. I won't try to do any better here, as I'm likely to make a subtle but important mistake.
The rub of it is that you can define "remainder" or "modulus" in different ways, and different languages have chosen different options to implement.
Jon Skeet
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I think your original answer struck right at the heart it. Sometimes '%' means "remainder" and sometimes it means "modulus", which aren't the same thing. – Bill the Lizard Jan 16 '09 at 13:56
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@Bill: I wasn't sure about deleting it, but then it comes down to the definition of "remainder" and "modulus" which are *also* (apparently) not as clear cut as one would like. – Jon Skeet Jan 16 '09 at 14:53
4
After dividing a number and a divisor, one of which is negative, you have at least two ways to separate them into a quotient and a remainder, such that quotient * divisor + remainder = number: you can either round the quotient towards negative infinity, or towards zero.
Many languages just choose one.
I can't resist pointing out that Common Lisp offers both.
Svante
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