ldiv.c 2.12 KB
Newer Older
Alexander Yurtsev's avatar
Alexander Yurtsev committed
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55
/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
   This file is part of the GNU C Library.

   The GNU C Library is free software; you can redistribute it and/or
   modify it under the terms of the GNU Library General Public License as
   published by the Free Software Foundation; either version 2 of the
   License, or (at your option) any later version.

   The GNU C Library is distributed in the hope that it will be useful,
   but WITHOUT ANY WARRANTY; without even the implied warranty of
   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
   Library General Public License for more details.

   You should have received a copy of the GNU Library General Public
   License along with the GNU C Library; see the file COPYING.LIB.  If not,
   write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
   Boston, MA 02111-1307, USA.  */

typedef struct {
	long    quot;
	long    rem;
} ldiv_t;
/* Return the `ldiv_t' representation of NUMER over DENOM.  */
ldiv_t
ldiv (long int numer, long int denom)
{
  ldiv_t result;

  result.quot = numer / denom;
  result.rem = numer % denom;

  /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
     NUMER / DENOM is to be computed in infinite precision.  In
     other words, we should always truncate the quotient towards
     zero, never -infinity.  Machine division and remainer may
     work either way when one or both of NUMER or DENOM is
     negative.  If only one is negative and QUOT has been
     truncated towards -infinity, REM will have the same sign as
     DENOM and the opposite sign of NUMER; if both are negative
     and QUOT has been truncated towards -infinity, REM will be
     positive (will have the opposite sign of NUMER).  These are
     considered `wrong'.  If both are NUM and DENOM are positive,
     RESULT will always be positive.  This all boils down to: if
     NUMER >= 0, but REM < 0, we got the wrong answer.  In that
     case, to get the right answer, add 1 to QUOT and subtract
     DENOM from REM.  */

  if (numer >= 0 && result.rem < 0)
    {
      ++result.quot;
      result.rem -= denom;
    }

  return result;
}