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div.c revision 1.1
      1  1.1  cgd /*
      2  1.1  cgd  * Copyright (c) 1990 Regents of the University of California.
      3  1.1  cgd  * All rights reserved.
      4  1.1  cgd  *
      5  1.1  cgd  * This code is derived from software contributed to Berkeley by
      6  1.1  cgd  * Chris Torek.
      7  1.1  cgd  *
      8  1.1  cgd  * Redistribution and use in source and binary forms, with or without
      9  1.1  cgd  * modification, are permitted provided that the following conditions
     10  1.1  cgd  * are met:
     11  1.1  cgd  * 1. Redistributions of source code must retain the above copyright
     12  1.1  cgd  *    notice, this list of conditions and the following disclaimer.
     13  1.1  cgd  * 2. Redistributions in binary form must reproduce the above copyright
     14  1.1  cgd  *    notice, this list of conditions and the following disclaimer in the
     15  1.1  cgd  *    documentation and/or other materials provided with the distribution.
     16  1.1  cgd  * 3. All advertising materials mentioning features or use of this software
     17  1.1  cgd  *    must display the following acknowledgement:
     18  1.1  cgd  *	This product includes software developed by the University of
     19  1.1  cgd  *	California, Berkeley and its contributors.
     20  1.1  cgd  * 4. Neither the name of the University nor the names of its contributors
     21  1.1  cgd  *    may be used to endorse or promote products derived from this software
     22  1.1  cgd  *    without specific prior written permission.
     23  1.1  cgd  *
     24  1.1  cgd  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
     25  1.1  cgd  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
     26  1.1  cgd  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
     27  1.1  cgd  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
     28  1.1  cgd  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
     29  1.1  cgd  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
     30  1.1  cgd  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
     31  1.1  cgd  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
     32  1.1  cgd  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
     33  1.1  cgd  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
     34  1.1  cgd  * SUCH DAMAGE.
     35  1.1  cgd  */
     36  1.1  cgd 
     37  1.1  cgd #if defined(LIBC_SCCS) && !defined(lint)
     38  1.1  cgd static char sccsid[] = "@(#)div.c	5.2 (Berkeley) 4/16/91";
     39  1.1  cgd #endif /* LIBC_SCCS and not lint */
     40  1.1  cgd 
     41  1.1  cgd #include <stdlib.h>		/* div_t */
     42  1.1  cgd 
     43  1.1  cgd div_t
     44  1.1  cgd div(num, denom)
     45  1.1  cgd 	int num, denom;
     46  1.1  cgd {
     47  1.1  cgd 	div_t r;
     48  1.1  cgd 
     49  1.1  cgd 	r.quot = num / denom;
     50  1.1  cgd 	r.rem = num % denom;
     51  1.1  cgd 	/*
     52  1.1  cgd 	 * The ANSI standard says that |r.quot| <= |n/d|, where
     53  1.1  cgd 	 * n/d is to be computed in infinite precision.  In other
     54  1.1  cgd 	 * words, we should always truncate the quotient towards
     55  1.1  cgd 	 * 0, never -infinity.
     56  1.1  cgd 	 *
     57  1.1  cgd 	 * Machine division and remainer may work either way when
     58  1.1  cgd 	 * one or both of n or d is negative.  If only one is
     59  1.1  cgd 	 * negative and r.quot has been truncated towards -inf,
     60  1.1  cgd 	 * r.rem will have the same sign as denom and the opposite
     61  1.1  cgd 	 * sign of num; if both are negative and r.quot has been
     62  1.1  cgd 	 * truncated towards -inf, r.rem will be positive (will
     63  1.1  cgd 	 * have the opposite sign of num).  These are considered
     64  1.1  cgd 	 * `wrong'.
     65  1.1  cgd 	 *
     66  1.1  cgd 	 * If both are num and denom are positive, r will always
     67  1.1  cgd 	 * be positive.
     68  1.1  cgd 	 *
     69  1.1  cgd 	 * This all boils down to:
     70  1.1  cgd 	 *	if num >= 0, but r.rem < 0, we got the wrong answer.
     71  1.1  cgd 	 * In that case, to get the right answer, add 1 to r.quot and
     72  1.1  cgd 	 * subtract denom from r.rem.
     73  1.1  cgd 	 */
     74  1.1  cgd 	if (num >= 0 && r.rem < 0) {
     75  1.1  cgd 		r.quot++;
     76  1.1  cgd 		r.rem -= denom;
     77  1.1  cgd 	}
     78  1.1  cgd 	return (r);
     79  1.1  cgd }
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