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Re: mod/div (was: reduce metaoperator on an empty list)

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From:
Mark Reed
Date:
May 23, 2005 15:54
Subject:
Re: mod/div (was: reduce metaoperator on an empty list)
Message ID:
BEB7D74A.609D%mark.reed@turner.com
I would really like to see ($x div $y) be (floor($x/$y)) and ($x mod $y) be
($x - $x div $y).   If the divisor is positive the modulus should be
positive, no matter what the sign of the dividend.  Avoids lots of special
case code across 0 boundaries.


On 2005-05-23 18:49, "TSa (Thomas SandlaƟ)" <Thomas.Sandlass@orthogon.com>
wrote:

> mark.a.biggar@comcast.net wrote:
>> > There are actuall two usefull definition for %.  The first which Ada calls
>> 'mod' always returns a value 0<=X<N and yes it has no working value that is
>> an identity.  The other which Ada calls 'rem' defined as follows:
> 
>> > 
>> > Signed integer division and remainder are defined by the relation:
>> > 
>> > A = (A/B)*B + (A rem B)
>> > 
>> >    where (A rem B) has the sign of A and an absolute value less than the
>> absolute value of B. Signed integer division satisfies the identity:
> 
>> > 
>> > (-A)/B = -(A/B) = A/(-B)
>> > 
>> > It does have a right side identity of +INF.
> 
> This is the truncating div-dominant definition of modulo.
> The eulerian definition is mod-dominant and nicely handles
> non-integer values. E.g.
> 
>   3.2 == 1.5 * 2 + 0.2 -+-->  3.2 / 1.5 == 2 + 0.2 / 1.5 == 2 + 1/15
>                         |               == 2 + 0.1333...
>                         +-->  3.2 % 1.5 == 0.2
> 
> Note that -3.2 == -4 + 0.8 == -4.5 + 1.3 == ...
> 
>            -3.2 / 1.5 == -3 + 1.3 / 1.5 == -3 + 0.8666... == -2.1333
>            -3.2 % 1.5 ==  1.3
> 
> With integers: 
> 
>       8  /   3  ==  (2 + 2/3) ==  2
>       8  / (-3) == -(2 + 2/3) == -2
>     (-8) /   3  == -(3 - 1/3) == -3  # this might surprise some people ;)
>     (-8) / (-3) ==  (3 - 1/3) ==  3
> 
>       8  % (-3) ==   8  % 3 == 2
>     (-8) % (-3) == (-8) % 3 == 1  # this as well, but it's just -3 * 3 + 1
> 
> Real valued division can be considered as % 0, that is infinite precision.
> While integer arithmetic is % 1. I.e. int $x == $x - $x % 1.
> 
>     floor $x == $x - $x % 1    # -1.2 - (-1.2) % 1 == -1.2 - 0.8 == -2
>     ceil  $x == 1 + floor $x
>     round $x == floor( $x + 0.5 )
>     trunc $x == $x < 0 ?? ceil $x :: floor $x
> 
> To @Larry: how are mod and div defined in Perl6?



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