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Newbie question: modulo operator with negative operand, bug or feature?

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Top Forums Shell Programming and Scripting Newbie question: modulo operator with negative operand, bug or feature?
# 1  
Old 08-11-2015
Newbie question: modulo operator with negative operand, bug or feature?
Hi,

I'm new to the Ash shell so my apologies if this is well known. In normal maths and other shells and languages I've used, the modulo operator always returns a positive remainder. For example see this discussion (first post so I can't hyperlink it):

http://math.stackexchange.com/questions/519845/modulo-of-a-negative-number

The Ash shell (and I am told Bash too) return a negative result. This code run on BusyBox illustrates the problem:

Code:
#!/bin/sh
echo "The result of A % B should be an integer in the range 0 to B-1"
echo "(9-7) % 5 = $(((9-7)%5)) (should be 2)."
echo "(7-9) % 5 = $(((7-9)%5)) (should be 3)."
exit 0

Is this accepted as just a 'feature' of the shells or is it a bug?
This User Gave Thanks to FleetFoot For This Post:
Scrutinizer
# 2  
Old 08-11-2015
Code:
echo "(9-7) % 5 = $(( $(( 9 - 7 )) % 5 )) (should be 2)."
# and
echo "(9-7) % 5 = $(( ( 9 - 7 ) % 5 )) (should be 2)."

Note the spaces.

Both work in bash. /bin/sh does not have to always equal bash, it should a modern POSIX shell. I do not know how /bin/sh evaluates on busybox.
# 3  
Old 08-11-2015
I tried every shell I could find and the answer was:
Code:
$ dash modtest
The result of A % B should be an integer in the range 0 to B-1
(9-7) % 5 = 2 (should be 2).
(7-9) % 5 = -2 (should be 3).

across all POSIX shells, dash (Debian Almquist Shell), bash, ksh88, ksh93, zsh...
# 4  
Old 08-11-2015
Code:
aia@localhost fleefoot]$ echo "(7-9) % 5 = $(((7-9)%5)) (should be 3)."
(7-9) % 5 = -2 (should be 3).

-2%5 is -2 according how the sign is handled in languages that satisfy it as a = ((a / b) * b) + (a % b)


An example to show the difference using Perl
Code:
[aia@localhost fleefoot]$ perl -le 'print -2%5'
3
[aia@localhost fleefoot]$ perl -le 'print (-2)%5'
-2

Last edited by Aia; 08-11-2015 at 02:51 PM..
# 5  
Old 08-11-2015
Hi.

Quote:
In computing, the modulo operation finds the remainder after division of one number by another (sometimes called modulus).

Given two positive numbers, a (the dividend) and n (the divisor), a modulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n.
..
When either a or n is negative, the naive definition breaks down and programming languages differ in how these values are defined.
-- https://en.wikipedia.org/wiki/Modulo_operation

Best wishes ... cheers, drl
These 2 Users Gave Thanks to drl For This Post:
Scrutinizer wisecracker
# 6  
Old 08-11-2015
@Scrutinizer - what did you try, the original code? The '$((' '))' construct requires spaces around it I believe.

NO! it does not need spaces, just leading spaces

I'm wrong - the $(( does not require spaces it is part of command substitution (from opnegroup.org):
Quote:
If the current character is an unquoted '$' or '`', the shell shall identify the start of any candidates for parameter expansion ( Parameter Expansion), command substitution ( Command Substitution), or arithmetic expansion ( Arithmetic Expansion) from their introductory unquoted character sequences: '$' or "${", "$(" or '`', and "$((", respectively.
But since that is true then why do we get wacky (divergent) results with POSIX-compliant shells? my ksh did not like it. My bash (3.4.3) did after I added some spaces. It specifically complained about %5. Hmm.
# 7  
Old 08-11-2015
Quote:
Originally Posted by Aia
[...]

-2%5 is -2 according how the sign is handled in languages that satisfy it as a = ((a / b) * b) + (a % b)


An example to show the difference using Perl
Code:
[aia@localhost fleefoot]$ perl -le 'print -2%5'
3
[aia@localhost fleefoot]$ perl -le 'print (-2)%5'
-2

Aia, could you elaborate on why that should render a different outcome ?

In awk:
Code:
awk 'BEGIN {print (-2)%5, -2%5}'
-2 -2

Quote:
Originally Posted by jim mcnamara
@Scrutinizer - what did you try, the original code? [..]
Hi Jim, yes, the original code...
Quote:

NO! it does not need spaces, just leading spaces

I'm wrong - the $(( does not require spaces it is part of command substitution (from opnegroup.org):


But since that is true then why do we get wacky (divergent) results with POSIX-compliant shells? my ksh did not like it. My bash (3.4.3) did after I added some spaces. It specifically complained about %5. Hmm.
What ksh and bash are those on what OS? Any ksh I tried and any bash on any OS did not seem to have a problem with it,
Last edited by Scrutinizer; 08-11-2015 at 01:45 PM..
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