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/*

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);

}