From: Rudolf Polzer <denshimeiru-sapmctacher@...>
Date: 2003-07-30T11:24:37+09:00
Subject: Re: -3 / 10 = -1  ?!

Scripsit ille �John Andrews� <john_b_andrews@yahoo.com>:
> Why does this program:
> 
>    print -3 / 10, "\n"
> 
> print -1?
> 
> If I underflow a negative int in C, I get zero.
> 
> If I underflow a negative integer in Perl (using "use integer"),
> I get zero.
> 
> Why does Ruby give me -1?

It's doing the "mathematical" division.

Integer division with remainder is mathematically defined as follows:

  For a given a \in Z and b \in N without 0 you calculate q \in Z
  and r \in [0, b[ with

    a = bq + r

  q is the quotient (in Ruby a/b) and r is the remainder (in Ruby a%b).
  I'll write for this: div(a, b) = (q, r).

Perl also does the "mathematical" division:

  rpolzer@katsuragi rpolzer $ perl -e 'print -1%10, "\n"'
  9

But:

  rpolzer@katsuragi rpolzer $ perl -Minteger -e 'print -1%10, "\n"'
  -1

as defined by Intel on x86 systems (other architectures may behave
differently). Without 'use integer', there's no such thing as integer
division in Perl, so I assume you were using it.

BTW: Ruby seems to define division by negative divisors like this:

                , r == 0: (-q, r)
  div(a, -b) = <
                ` r != 0: (-q-1, r-b)

or like above if you define [0, b[ being equivalent to ]b, 0] for
negative b.


-- 
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>[<<->>[-]]<<[->[<<<+>>>-]<]>]<+<<[>>-<<[-]]>>[-<[>+<<<<.>>>-]>[<+>-]<<<.>>>]<],