Return to div.c CVS log

Up to [local] / prex-old / usr / lib / libc / stdlib

Annotation of prex-old/usr/lib/libc/stdlib/div.c, Revision 1.1.1.1

1.1       nbrk        1: /*
                      2:  * Copyright (c) 1990, 1993
                      3:  *     The Regents of the University of California.  All rights reserved.
                      4:  *
                      5:  * This code is derived from software contributed to Berkeley by
                      6:  * Chris Torek.
                      7:  *
                      8:  * Redistribution and use in source and binary forms, with or without
                      9:  * modification, are permitted provided that the following conditions
                     10:  * are met:
                     11:  * 1. Redistributions of source code must retain the above copyright
                     12:  *    notice, this list of conditions and the following disclaimer.
                     13:  * 2. Redistributions in binary form must reproduce the above copyright
                     14:  *    notice, this list of conditions and the following disclaimer in the
                     15:  *    documentation and/or other materials provided with the distribution.
                     16:  * 3. Neither the name of the University nor the names of its contributors
                     17:  *    may be used to endorse or promote products derived from this software
                     18:  *    without specific prior written permission.
                     19:  *
                     20:  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
                     21:  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
                     22:  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
                     23:  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
                     24:  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
                     25:  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
                     26:  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
                     27:  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
                     28:  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
                     29:  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
                     30:  * SUCH DAMAGE.
                     31:  */
                     32:
                     33: #include <stdlib.h>            /* div_t */
                     34:
                     35: div_t
                     36: div(num, denom)
                     37:        int num, denom;
                     38: {
                     39:        div_t r;
                     40:
                     41:        r.quot = num / denom;
                     42:        r.rem = num % denom;
                     43:        /*
                     44:         * The ANSI standard says that |r.quot| <= |n/d|, where
                     45:         * n/d is to be computed in infinite precision.  In other
                     46:         * words, we should always truncate the quotient towards
                     47:         * 0, never -infinity.
                     48:         *
                     49:         * Machine division and remainer may work either way when
                     50:         * one or both of n or d is negative.  If only one is
                     51:         * negative and r.quot has been truncated towards -inf,
                     52:         * r.rem will have the same sign as denom and the opposite
                     53:         * sign of num; if both are negative and r.quot has been
                     54:         * truncated towards -inf, r.rem will be positive (will
                     55:         * have the opposite sign of num).  These are considered
                     56:         * `wrong'.
                     57:         *
                     58:         * If both are num and denom are positive, r will always
                     59:         * be positive.
                     60:         *
                     61:         * This all boils down to:
                     62:         *      if num >= 0, but r.rem < 0, we got the wrong answer.
                     63:         * In that case, to get the right answer, add 1 to r.quot and
                     64:         * subtract denom from r.rem.
                     65:         */
                     66:        if (num >= 0 && r.rem < 0) {
                     67:                r.quot++;
                     68:                r.rem -= denom;
                     69:        }
                     70:        return (r);
                     71: }