genericopenlibs/cstdlib/LSTDLIB/DIV.C
author Peter Fordham <peter.fordham@gmail.com>
Mon, 22 Mar 2010 12:57:04 -0700
branchCompilerCompatibility
changeset 13 ef48f5dc1b7f
parent 0 e4d67989cc36
permissions -rw-r--r--
Bug 1713 - You can't check the existence of a typedef using the preprocessor. Added a preprocessor check. Also removed an overqualified method name.

/*
* 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]>;
* Divide
* @tex
* $n/d$,
* @end tex
* @ifinfo
* <[n]>/<[d]>,
* @end ifinfo
* 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 */

/**
Divide two integer values.
numer is divided by denom. 
Quotient and remainder are returned in a div_t structure.
@return   A div_t structure is returned
@param num Numerator. 
@param denom Denominator.
*/
EXPORT_C div_t div (int num, 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.
	 *
	 * 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 && r.rem < 0) {
		r.quot++;
		r.rem -= denom;
	}
	return (r);
}