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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);
}
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