stdlib/ldiv.c

   1 /* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
   2    This file is part of the GNU C Library.
   3
   4    The GNU C Library is free software; you can redistribute it and/or
   5    modify it under the terms of the GNU Lesser General Public
   6    License as published by the Free Software Foundation; either
   7    version 2.1 of the License, or (at your option) any later version.
   8
   9    The GNU C Library is distributed in the hope that it will be useful,
  10    but WITHOUT ANY WARRANTY; without even the implied warranty of
  11    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  12    Lesser General Public License for more details.
  13
  14    You should have received a copy of the GNU Lesser General Public
  15    License along with the GNU C Library; if not, write to the Free
  16    Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
  17    02111-1307 USA.  */
  18
  19 #include <stdlib.h>
  20
  21
  22 /* Return the `ldiv_t' representation of NUMER over DENOM.  */
  23 ldiv_t
  24 ldiv (long int numer, long int denom)
  25 {
  26   ldiv_t result;
  27
  28   result.quot = numer / denom;
  29   result.rem = numer % denom;
  30
  31   /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
  32      NUMER / DENOM is to be computed in infinite precision.  In
  33      other words, we should always truncate the quotient towards
  34      zero, never -infinity.  Machine division and remainer may
  35      work either way when one or both of NUMER or DENOM is
  36      negative.  If only one is negative and QUOT has been
  37      truncated towards -infinity, REM will have the same sign as
  38      DENOM and the opposite sign of NUMER; if both are negative
  39      and QUOT has been truncated towards -infinity, REM will be
  40      positive (will have the opposite sign of NUMER).  These are
  41      considered `wrong'.  If both are NUM and DENOM are positive,
  42      RESULT will always be positive.  This all boils down to: if
  43      NUMER >= 0, but REM < 0, we got the wrong answer.  In that
  44      case, to get the right answer, add 1 to QUOT and subtract
  45      DENOM from REM.  */
  46
  47   if (numer >= 0 && result.rem < 0)
  48     {
  49       ++result.quot;
  50       result.rem -= denom;
  51     }
  52
  53   return result;
  54 }