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133 lines
4.1 KiB
C
133 lines
4.1 KiB
C
/*
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FUNCTION
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<<div>>---divide two integers
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INDEX
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div
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ANSI_SYNOPSIS
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#include <stdlib.h>
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div_t div(int <[n]>, int <[d]>);
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TRAD_SYNOPSIS
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#include <stdlib.h>
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div_t div(<[n]>, <[d]>)
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int <[n]>, <[d]>;
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DESCRIPTION
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Divide
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@tex
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$n/d$,
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@end tex
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@ifnottex
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<[n]>/<[d]>,
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@end ifnottex
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returning quotient and remainder as two integers in a structure <<div_t>>.
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RETURNS
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The result is represented with the structure
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. typedef struct
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. {
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. int quot;
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. int rem;
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. } div_t;
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where the <<quot>> field represents the quotient, and <<rem>> the
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remainder. For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then
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<[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'.
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To divide <<long>> rather than <<int>> values, use the similar
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function <<ldiv>>.
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PORTABILITY
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<<div>> is ANSI.
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No supporting OS subroutines are required.
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*/
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/*
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* Copyright (c) 1990 Regents of the University of California.
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* All rights reserved.
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*
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* This code is derived from software contributed to Berkeley by
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* Chris Torek.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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* 3. All advertising materials mentioning features or use of this software
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* must display the following acknowledgement:
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* This product includes software developed by the University of
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* California, Berkeley and its contributors.
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* 4. Neither the name of the University nor the names of its contributors
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* may be used to endorse or promote products derived from this software
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* without specific prior written permission.
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*
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* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*/
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#include <_ansi.h>
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#include <stdlib.h> /* div_t */
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div_t
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_DEFUN (div, (num, denom),
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int num _AND
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int denom)
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{
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div_t r;
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r.quot = num / denom;
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r.rem = num % denom;
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/*
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* The ANSI standard says that |r.quot| <= |n/d|, where
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* n/d is to be computed in infinite precision. In other
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* words, we should always truncate the quotient towards
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* 0, never -infinity or +infinity.
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*
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* Machine division and remainer may work either way when
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* one or both of n or d is negative. If only one is
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* negative and r.quot has been truncated towards -inf,
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* r.rem will have the same sign as denom and the opposite
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* sign of num; if both are negative and r.quot has been
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* truncated towards -inf, r.rem will be positive (will
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* have the opposite sign of num). These are considered
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* `wrong'.
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*
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* If both are num and denom are positive, r will always
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* be positive.
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*
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* This all boils down to:
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* if num >= 0, but r.rem < 0, we got the wrong answer.
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* In that case, to get the right answer, add 1 to r.quot and
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* subtract denom from r.rem.
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* if num < 0, but r.rem > 0, we also have the wrong answer.
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* In this case, to get the right answer, subtract 1 from r.quot and
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* add denom to r.rem.
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*/
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if (num >= 0 && r.rem < 0) {
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++r.quot;
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r.rem -= denom;
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}
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else if (num < 0 && r.rem > 0) {
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--r.quot;
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r.rem += denom;
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}
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return (r);
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}
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