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[/] [openrisc/] [trunk/] [gnu-src/] [newlib-1.17.0/] [newlib/] [libc/] [stdlib/] [div.c] - Blame information for rev 158

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1 148 jeremybenn
/*
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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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