Maintain runtime of each benchmark at ~10 seconds
[glibc.git] / stdlib / div.c
blob44a30a7ea485a9eef38fb4ffe6aaebdc06468b3b
1 /* Copyright (C) 1992-2013 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, see
16 <http://www.gnu.org/licenses/>. */
19 * Copyright (c) 1990 Regents of the University of California.
20 * All rights reserved.
22 * This code is derived from software contributed to Berkeley by
23 * Chris Torek.
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26 * modification, are permitted provided that the following conditions
27 * are met:
28 * 1. Redistributions of source code must retain the above copyright
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44 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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47 * SUCH DAMAGE.
50 #include <stdlib.h>
52 /* Return the `div_t' representation of NUMER over DENOM. */
53 div_t
54 div (numer, denom)
55 int numer, denom;
57 div_t result;
59 result.quot = numer / denom;
60 result.rem = numer % denom;
62 /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
63 NUMER / DENOM is to be computed in infinite precision. In
64 other words, we should always truncate the quotient towards
65 zero, never -infinity. Machine division and remainer may
66 work either way when one or both of NUMER or DENOM is
67 negative. If only one is negative and QUOT has been
68 truncated towards -infinity, REM will have the same sign as
69 DENOM and the opposite sign of NUMER; if both are negative
70 and QUOT has been truncated towards -infinity, REM will be
71 positive (will have the opposite sign of NUMER). These are
72 considered `wrong'. If both are NUM and DENOM are positive,
73 RESULT will always be positive. This all boils down to: if
74 NUMER >= 0, but REM < 0, we got the wrong answer. In that
75 case, to get the right answer, add 1 to QUOT and subtract
76 DENOM from REM. */
78 if (numer >= 0 && result.rem < 0)
80 ++result.quot;
81 result.rem -= denom;
84 return result;