55 lines
2.1 KiB
C
55 lines
2.1 KiB
C
/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Library General Public License as
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published by the Free Software Foundation; either version 2 of the
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License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Library General Public License for more details.
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You should have received a copy of the GNU Library General Public
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License along with the GNU C Library; see the file COPYING.LIB. If not,
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write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
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Boston, MA 02111-1307, USA. */
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typedef struct {
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long quot;
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long rem;
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} ldiv_t;
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/* Return the `ldiv_t' representation of NUMER over DENOM. */
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ldiv_t
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ldiv (long int numer, long int denom)
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{
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ldiv_t result;
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result.quot = numer / denom;
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result.rem = numer % denom;
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/* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
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NUMER / DENOM is to be computed in infinite precision. In
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other words, we should always truncate the quotient towards
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zero, never -infinity. Machine division and remainer may
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work either way when one or both of NUMER or DENOM is
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negative. If only one is negative and QUOT has been
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truncated towards -infinity, REM will have the same sign as
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DENOM and the opposite sign of NUMER; if both are negative
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and QUOT has been truncated towards -infinity, REM will be
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positive (will have the opposite sign of NUMER). These are
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considered `wrong'. If both are NUM and DENOM are positive,
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RESULT will always be positive. This all boils down to: if
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NUMER >= 0, but REM < 0, we got the wrong answer. In that
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case, to get the right answer, add 1 to QUOT and subtract
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DENOM from REM. */
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if (numer >= 0 && result.rem < 0)
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{
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++result.quot;
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result.rem -= denom;
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}
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return result;
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}
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