Why is modulus different in different programming languages?

问题

Perl

print 2 % -18;

-->

-16

Tcl

puts [expr {2 % -18}]

-->

-16

but VBScript

wscript.echo 2 mod -18

-->

Why the difference?

回答1:

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). "

回答2:

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.

回答3:

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.