newlib/libc/stdlib/div.c

   1 /*
   2 FUNCTION
   3 <<div>>---divide two integers
   4
   5 INDEX
   6         div
   7
   8 SYNOPSIS
   9         #include <stdlib.h>
  10         div_t div(int <[n]>, int <[d]>);
  11
  12 DESCRIPTION
  13 Divide
  14 @tex
  15 $n/d$,
  16 @end tex
  17 @ifnottex
  18 <[n]>/<[d]>,
  19 @end ifnottex
  20 returning quotient and remainder as two integers in a structure <<div_t>>.
  21
  22 RETURNS
  23 The result is represented with the structure
  24
  25 . typedef struct
  26 . {
  27 .  int quot;
  28 .  int rem;
  29 . } div_t;
  30
  31 where the <<quot>> field represents the quotient, and <<rem>> the
  32 remainder.  For nonzero <[d]>, if `<<<[r]> = div(<[n]>,<[d]>);>>' then
  33 <[n]> equals `<<<[r]>.rem + <[d]>*<[r]>.quot>>'.
  34
  35 To divide <<long>> rather than <<int>> values, use the similar
  36 function <<ldiv>>.
  37
  38 PORTABILITY
  39 <<div>> is ANSI.
  40
  41 No supporting OS subroutines are required.
  42 */
  43
  44 /*
  45  * Copyright (c) 1990 Regents of the University of California.
  46  * All rights reserved.
  47  *
  48  * This code is derived from software contributed to Berkeley by
  49  * Chris Torek.
  50  *
  51  * Redistribution and use in source and binary forms, with or without
  52  * modification, are permitted provided that the following conditions
  53  * are met:
  54  * 1. Redistributions of source code must retain the above copyright
  55  *    notice, this list of conditions and the following disclaimer.
  56  * 2. Redistributions in binary form must reproduce the above copyright
  57  *    notice, this list of conditions and the following disclaimer in the
  58  *    documentation and/or other materials provided with the distribution.
  59  * 3. Neither the name of the University nor the names of its contributors
  60  *    may be used to endorse or promote products derived from this software
  61  *    without specific prior written permission.
  62  *
  63  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
  64  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
  65  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
  66  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
  67  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
  68  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
  69  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
  70  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
  71  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
  72  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
  73  * SUCH DAMAGE.
  74  */
  75
  76 #include <_ansi.h>
  77 #include <stdlib.h>             /* div_t */
  78
  79 div_t
  80 div (int num,
  81         int denom)
  82 {
  83         div_t r;
  84
  85         r.quot = num / denom;
  86         r.rem = num % denom;
  87         /*
  88          * The ANSI standard says that |r.quot| <= |n/d|, where
  89          * n/d is to be computed in infinite precision.  In other
  90          * words, we should always truncate the quotient towards
  91          * 0, never -infinity or +infinity.
  92          *
  93          * Machine division and remainer may work either way when
  94          * one or both of n or d is negative.  If only one is
  95          * negative and r.quot has been truncated towards -inf,
  96          * r.rem will have the same sign as denom and the opposite
  97          * sign of num; if both are negative and r.quot has been
  98          * truncated towards -inf, r.rem will be positive (will
  99          * have the opposite sign of num).  These are considered
 100          * `wrong'.
 101          *
 102          * If both are num and denom are positive, r will always
 103          * be positive.
 104          *
 105          * This all boils down to:
 106          *      if num >= 0, but r.rem < 0, we got the wrong answer.
 107          * In that case, to get the right answer, add 1 to r.quot and
 108          * subtract denom from r.rem.
 109          *      if num < 0, but r.rem > 0, we also have the wrong answer.
 110          * In this case, to get the right answer, subtract 1 from r.quot and
 111          * add denom to r.rem.
 112          */
 113         if (num >= 0 && r.rem < 0) {
 114                 ++r.quot;
 115                 r.rem -= denom;
 116         }
 117         else if (num < 0 && r.rem > 0) {
 118                 --r.quot;
 119                 r.rem += denom;
 120         }
 121         return (r);
 122 }