From: Artem Voroztsov <artem.voroztsov@...>
Date: 2008-03-04T19:06:56+09:00
Subject: Why not "truncation towards zero"?

Is it intentional that  (-1/2)  gives -1 ?

Please, give me links if there is any rationale.

(i have ruby 1.8.6 (2007-03-13 patchlevel 0) [i386-mswin32] )

# Ruby
puts -1/2; #  => -1

// C (
#include <stdio.h>
int main() {
  printf("%d\n", -1/2); // => 0 (in most platforms. )
  return 0;
}

It looks like  standard C9x  states not only  equation b*(a/b) + a%b == a
but also "truncation towards zero".


In http://www.open-std.org/JTC1/SC22/WG14/www/docs/C99RationaleV5.10.pdf
one  could read (6.5.5, page 74):
  .. In Fortran, however, the result will always
  truncate toward zero, and the overhead seems to be acceptable to the
numeric
  programming  community. Therefore, C99 now requires similar behavior, which
  should facilitate porting of code from Fortran to C...