lib/libc/gen/modf_ieee754.c

   1 /* $NetBSD: modf_ieee754.c,v 1.1 2003/05/12 15:15:16 kleink Exp $ */
   2
   3 /*
   4  * Copyright (c) 1994, 1995 Carnegie-Mellon University.
   5  * All rights reserved.
   6  *
   7  * Author: Chris G. Demetriou
   8  *
   9  * Permission to use, copy, modify and distribute this software and
  10  * its documentation is hereby granted, provided that both the copyright
  11  * notice and this permission notice appear in all copies of the
  12  * software, derivative works or modified versions, and any portions
  13  * thereof, and that both notices appear in supporting documentation.
  14  *
  15  * CARNEGIE MELLON ALLOWS FREE USE OF THIS SOFTWARE IN ITS "AS IS"
  16  * CONDITION.  CARNEGIE MELLON DISCLAIMS ANY LIABILITY OF ANY KIND
  17  * FOR ANY DAMAGES WHATSOEVER RESULTING FROM THE USE OF THIS SOFTWARE.
  18  *
  19  * Carnegie Mellon requests users of this software to return to
  20  *
  21  *  Software Distribution Coordinator  or  Software.Distribution@CS.CMU.EDU
  22  *  School of Computer Science
  23  *  Carnegie Mellon University
  24  *  Pittsburgh PA 15213-3890
  25  *
  26  * any improvements or extensions that they make and grant Carnegie the
  27  * rights to redistribute these changes.
  28  */
  29
  30 #include <sys/types.h>
  31 #include <machine/ieee.h>
  32 #include <errno.h>
  33 #include <math.h>
  34
  35 /*
  36  * double modf(double val, double *iptr)
  37  * returns: f and i such that |f| < 1.0, (f + i) = val, and
  38  *      sign(f) == sign(i) == sign(val).
  39  *
  40  * Beware signedness when doing subtraction, and also operand size!
  41  */
  42 double
  43 modf(double val, double *iptr)
  44 {
  45         union ieee_double_u u, v;
  46         u_int64_t frac;
  47
  48         /*
  49          * If input is Inf or NaN, return it and leave i alone.
  50          */
  51         u.dblu_d = val;
  52         if (u.dblu_dbl.dbl_exp == DBL_EXP_INFNAN)
  53                 return (u.dblu_d);
  54
  55         /*
  56          * If input can't have a fractional part, return
  57          * (appropriately signed) zero, and make i be the input.
  58          */
  59         if ((int)u.dblu_dbl.dbl_exp - DBL_EXP_BIAS > DBL_FRACBITS - 1) {
  60                 *iptr = u.dblu_d;
  61                 v.dblu_d = 0.0;
  62                 v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
  63                 return (v.dblu_d);
  64         }
  65
  66         /*
  67          * If |input| < 1.0, return it, and set i to the appropriately
  68          * signed zero.
  69          */
  70         if (u.dblu_dbl.dbl_exp < DBL_EXP_BIAS) {
  71                 v.dblu_d = 0.0;
  72                 v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
  73                 *iptr = v.dblu_d;
  74                 return (u.dblu_d);
  75         }
  76
  77         /*
  78          * There can be a fractional part of the input.
  79          * If you look at the math involved for a few seconds, it's
  80          * plain to see that the integral part is the input, with the
  81          * low (DBL_FRACBITS - (exponent - DBL_EXP_BIAS)) bits zeroed,
  82          * the fractional part is the part with the rest of the
  83          * bits zeroed.  Just zeroing the high bits to get the
  84          * fractional part would yield a fraction in need of
  85          * normalization.  Therefore, we take the easy way out, and
  86          * just use subtraction to get the fractional part.
  87          */
  88         v.dblu_d = u.dblu_d;
  89         /* Zero the low bits of the fraction, the sleazy way. */
  90         frac = ((u_int64_t)v.dblu_dbl.dbl_frach << 32) + v.dblu_dbl.dbl_fracl;
  91         frac >>= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
  92         frac <<= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
  93         v.dblu_dbl.dbl_fracl = frac & 0xffffffff;
  94         v.dblu_dbl.dbl_frach = frac >> 32;
  95         *iptr = v.dblu_d;
  96
  97         u.dblu_d -= v.dblu_d;
  98         u.dblu_dbl.dbl_sign = v.dblu_dbl.dbl_sign;
  99         return (u.dblu_d);
 100 }