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dc36d6f9bb
Remove ancient SCCS tags from the tree, automated scripting, with two minor fixup to keep things compiling. All the common forms in the tree were removed with a perl script. Sponsored by: Netflix
74 lines
2.8 KiB
C
74 lines
2.8 KiB
C
/*-
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* SPDX-License-Identifier: BSD-3-Clause
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*
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* Copyright (c) 1990, 1993
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* The Regents of the University of California. All rights reserved.
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*
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* This code is derived from software contributed to Berkeley by
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* Chris Torek.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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* 3. Neither the name of the University nor the names of its contributors
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* may be used to endorse or promote products derived from this software
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* without specific prior written permission.
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*
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* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*/
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#include <stdlib.h> /* div_t */
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div_t
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div(int num, int denom)
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{
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div_t r;
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r.quot = num / denom;
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r.rem = num % denom;
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#if !defined(__STDC_VERSION__) || (__STDC_VERSION__ < 199901L)
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/*
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* The ANSI standard says that |r.quot| <= |n/d|, where
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* n/d is to be computed in infinite precision. In other
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* words, we should always truncate the quotient towards
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* 0, never -infinity.
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*
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* Machine division and remainer may work either way when
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* one or both of n or d is negative. If only one is
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* negative and r.quot has been truncated towards -inf,
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* r.rem will have the same sign as denom and the opposite
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* sign of num; if both are negative and r.quot has been
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* truncated towards -inf, r.rem will be positive (will
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* have the opposite sign of num). These are considered
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* `wrong'.
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*
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* If both are num and denom are positive, r will always
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* be positive.
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*
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* This all boils down to:
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* if num >= 0, but r.rem < 0, we got the wrong answer.
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* In that case, to get the right answer, add 1 to r.quot and
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* subtract denom from r.rem.
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*/
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if (num >= 0 && r.rem < 0) {
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r.quot++;
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r.rem -= denom;
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}
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#endif
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return (r);
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}
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