compiler/stdc/div.c

   1 /*
   2     Copyright © 1995-2012, The AROS Development Team. All rights reserved.
   3     $Id$
   4
   5     C99 function div().
   6 */
   7
   8 /*****************************************************************************
   9
  10     NAME */
  11 #include <stdlib.h>
  12
  13         div_t div (
  14
  15 /*  SYNOPSIS */
  16         int numer,
  17         int denom)
  18
  19 /*  FUNCTION
  20         Compute quotient en remainder of two int variables
  21
  22     INPUTS
  23         numer = the numerator
  24         denom = the denominator
  25
  26     RESULT
  27         a struct with two ints quot and rem with
  28         quot = numer / denom and rem = numer % denom.
  29
  30         typedef struct div_t {
  31             int quot;
  32             int rem;
  33         } div_t;
  34
  35    NOTES
  36
  37     EXAMPLE
  38
  39     BUGS
  40
  41     SEE ALSO
  42         ldiv()
  43
  44     INTERNALS
  45
  46 ******************************************************************************/
  47 {
  48     div_t ret;
  49
  50     ret.quot = numer / denom;
  51     ret.rem  = numer % denom;
  52
  53     /*
  54         This comment is from FreeBSD's src/lib/libc/stdlib/div.c, but
  55         the code was written before the comment was added (see CVS log).
  56
  57         Thus the code isn't really under the BSD license. The comment is...
  58      */
  59
  60     /*
  61      * The ANSI standard says that |r.quot| <= |n/d|, where
  62      * n/d is to be computed in infinite precision.  In other
  63      * words, we should always truncate the quotient towards
  64      * 0, never -infinity.
  65      *
  66      * Machine division and remainer may work either way when
  67      * one or both of n or d is negative.  If only one is
  68      * negative and r.quot has been truncated towards -inf,
  69      * r.rem will have the same sign as denom and the opposite
  70      * sign of num; if both are negative and r.quot has been
  71      * truncated towards -inf, r.rem will be positive (will
  72      * have the opposite sign of num).  These are considered
  73      * `wrong'.
  74      *
  75      * If both are num and denom are positive, r will always
  76      * be positive.
  77      *
  78      * This all boils down to:
  79      *      if num >= 0, but r.rem < 0, we got the wrong answer.
  80      * In that case, to get the right answer, add 1 to r.quot and
  81      * subtract denom from r.rem.
  82      */
  83     if (numer >= 0 && ret.rem < 0)
  84     {
  85         ret.quot++;
  86         ret.rem -= denom;
  87     }
  88     return ret;
  89 }