Add bho_IN locale
[glibc.git] / stdlib / div.c
blob5268f4c49440c936e23049d7eda4fde68f650369
1 /* Copyright (C) 1992, 1997, 1999 Free Software Foundation, Inc.
2 This file is part of the GNU C Library.
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.
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.
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. */
20 * Copyright (c) 1990 Regents of the University of California.
21 * All rights reserved.
23 * This code is derived from software contributed to Berkeley by
24 * Chris Torek.
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27 * modification, are permitted provided that the following conditions
28 * are met:
29 * 1. Redistributions of source code must retain the above copyright
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48 * SUCH DAMAGE.
51 #include <stdlib.h>
53 /* Return the `div_t' representation of NUMER over DENOM. */
54 div_t
55 div (numer, denom)
56 int numer, denom;
58 div_t result;
60 result.quot = numer / denom;
61 result.rem = numer % denom;
63 /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
64 NUMER / DENOM is to be computed in infinite precision. In
65 other words, we should always truncate the quotient towards
66 zero, never -infinity. Machine division and remainer may
67 work either way when one or both of NUMER or DENOM is
68 negative. If only one is negative and QUOT has been
69 truncated towards -infinity, REM will have the same sign as
70 DENOM and the opposite sign of NUMER; if both are negative
71 and QUOT has been truncated towards -infinity, REM will be
72 positive (will have the opposite sign of NUMER). These are
73 considered `wrong'. If both are NUM and DENOM are positive,
74 RESULT will always be positive. This all boils down to: if
75 NUMER >= 0, but REM < 0, we got the wrong answer. In that
76 case, to get the right answer, add 1 to QUOT and subtract
77 DENOM from REM. */
79 if (numer >= 0 && result.rem < 0)
81 ++result.quot;
82 result.rem -= denom;
85 return result;