stdlib/lldiv.c

   1 /* `long long int' divison with remainder.
   2    Copyright (C) 1992-2013 Free Software Foundation, Inc.
   3    This file is part of the GNU C Library.
   4
   5    The GNU C Library is free software; you can redistribute it and/or
   6    modify it under the terms of the GNU Lesser General Public
   7    License as published by the Free Software Foundation; either
   8    version 2.1 of the License, or (at your option) any later version.
   9
  10    The GNU C Library is distributed in the hope that it will be useful,
  11    but WITHOUT ANY WARRANTY; without even the implied warranty of
  12    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  13    Lesser General Public License for more details.
  14
  15    You should have received a copy of the GNU Lesser General Public
  16    License along with the GNU C Library; if not, see
  17    <http://www.gnu.org/licenses/>.  */
  18
  19 #include <stdlib.h>
  20
  21
  22 /* Return the `lldiv_t' representation of NUMER over DENOM.  */
  23 lldiv_t
  24 lldiv (numer, denom)
  25      long long int numer;
  26      long long int denom;
  27 {
  28   lldiv_t result;
  29
  30   result.quot = numer / denom;
  31   result.rem = numer % denom;
  32
  33   /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
  34      NUMER / DENOM is to be computed in infinite precision.  In
  35      other words, we should always truncate the quotient towards
  36      zero, never -infinity.  Machine division and remainer may
  37      work either way when one or both of NUMER or DENOM is
  38      negative.  If only one is negative and QUOT has been
  39      truncated towards -infinity, REM will have the same sign as
  40      DENOM and the opposite sign of NUMER; if both are negative
  41      and QUOT has been truncated towards -infinity, REM will be
  42      positive (will have the opposite sign of NUMER).  These are
  43      considered `wrong'.  If both are NUM and DENOM are positive,
  44      RESULT will always be positive.  This all boils down to: if
  45      NUMER >= 0, but REM < 0, we got the wrong answer.  In that
  46      case, to get the right answer, add 1 to QUOT and subtract
  47      DENOM from REM.  */
  48
  49   if (numer >= 0 && result.rem < 0)
  50     {
  51       ++result.quot;
  52       result.rem -= denom;
  53     }
  54
  55   return result;
  56 }