Re: [UUG/MLUG] perl and y2k

["Christopher J. Kaiser" <EMAIL:PROTECTED>]

Tymm Twillman wrote:
> 
> you know, this then caused me to go look at perlop(1); description there:
> 
> Binary ``%'' computes the modulus of two numbers. Given integer operands
> $a and $b: If $b is positive, then $a % $b is $a minus the largest
> multiple of $b that is not greater
> than $a. 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).
> 
> (got that?)
> 
> which means that perl is broken as re: documentation.  Probably so many
> people expect modulus to always return a positive # that they broke it.
> 
> -Tymm

Nope.  As per above: $a = -98, $b = 100.  Since $b is positive, rule one
applies.  Therefore,  $a % $b = %a - $b * x, where $b * x < $a.  Thus, x
= -2, $b * x = -200, and $a % $b = -98 - (-200) = 2.

Looks like what perl returned for me.

Chris
-- 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~       
Christopher J. Kaiser  
Assistant Professor
Animal Sciences Unit
University of Missouri
Columbia, MO 65211
EMAIL:PROTECTED
phone: 573.882.5479
fax: 573.884.4606
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