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/* `long long int' divison with remainder.
Copyright (C) 1992, 1996 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. */
#include <stdlib.h>
/* Return the `lldiv_t' representation of NUMER over DENOM. */
lldiv_t
lldiv (numer, denom)
long long int numer;
long long int denom;
{
lldiv_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;
}
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