usr/src/lib/libc/stdlib/div.c

/*
 * Copyright (c) 1990 Regents of the University of California.
 * All rights reserved.
 *
 * This code is derived from software contributed to Berkeley by
 * Chris Torek.
 *
 * %sccs.include.redist.c%
 */

#if defined(LIBC_SCCS) && !defined(lint)
static char sccsid[] = "@(#)div.c       5.2 (Berkeley) %G%";
#endif /* LIBC_SCCS and not lint */

#include <stdlib.h>             /* div_t */

div_t
div(num, denom)
        int num, 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);
}