div.c

Newlib 嵌入式 C库 标准实现代码

/*
FUNCTION
<<div>>---divide two integers

INDEX
	div

ANSI_SYNOPSIS
	#include <stdlib.h>
	div_t div(int <[n]>, int <[d]>);

TRAD_SYNOPSIS
	#include <stdlib.h>
	div_t div(<[n]>, <[d]>)
	int <[n]>, <[d]>;

DESCRIPTION
Divide
@tex
$n/d$,
@end tex
@ifnottex
<[n]>/<[d]>,
@end ifnottex
returning quotient and remainder as two integers in a structure <<div_t>>.

RETURNS
The result is represented with the structure

. typedef struct
. {
.  int quot;
.  int rem;
. } div_t;

where the <<quot>> field represents the quotient, and <<rem>> the
remainder.  For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then
<[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'.

To divide <<long>> rather than <<int>> values, use the similar
function <<ldiv>>.

PORTABILITY
<<div>> is ANSI.

No supporting OS subroutines are required.
*/

/*
 * Copyright (c) 1990 Regents of the University of California.
 * All rights reserved.
 *
 * This code is derived from software contributed to Berkeley by
 * Chris Torek.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 * 3. All advertising materials mentioning features or use of this software
 *    must display the following acknowledgement:
 *	This product includes software developed by the University of
 *	California, Berkeley and its contributors.
 * 4. Neither the name of the University nor the names of its contributors
 *    may be used to endorse or promote products derived from this software
 *    without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
 * SUCH DAMAGE.
 */

#include <_ansi.h>
#include <stdlib.h>		/* div_t */

div_t
_DEFUN (div, (num, denom),
	int num _AND
	int denom)
{
	div_t r;

	r.quot = num / denom;
	r.rem = num % denom;
	/*
	 * The ANSI standard says that |r.quot| <= |n/d|, where
	 * n/d is to be computed in infinite precision.  In other
	 * words, we should always truncate the quotient towards
	 * 0, never -infinity or +infinity.
	 *
	 * Machine division and remainer may work either way when
	 * one or both of n or d is negative.  If only one is
	 * negative and r.quot has been truncated towards -inf,
	 * r.rem will have the same sign as denom and the opposite
	 * sign of num; if both are negative and r.quot has been
	 * truncated towards -inf, r.rem will be positive (will
	 * have the opposite sign of num).  These are considered
	 * `wrong'.
	 *
	 * If both are num and denom are positive, r will always
	 * be positive.
	 *
	 * This all boils down to:
	 *	if num >= 0, but r.rem < 0, we got the wrong answer.
	 * In that case, to get the right answer, add 1 to r.quot and
	 * subtract denom from r.rem.
	 *      if num < 0, but r.rem > 0, we also have the wrong answer.
	 * In this case, to get the right answer, subtract 1 from r.quot and
	 * add denom to r.rem.
	 */
	if (num >= 0 && r.rem < 0) {
		++r.quot;
		r.rem -= denom;
	}
	else if (num < 0 && r.rem > 0) {
		--r.quot;
		r.rem += denom;
	}
	return (r);
}