5261 libm should stop using synonyms.h
5298 fabs is 0-sized, confuses dis(1) and others
Reviewed by: Josef 'Jeff' Sipek <jeffpc@josefsipek.net>
Approved by: Gordon Ross <gwr@nexenta.com>

          --- old/usr/src/lib/libm/common/C/exp10.c
          +++ new/usr/src/lib/libm/common/C/exp10.c

   1    1  /*
   2    2   * CDDL HEADER START
   3    3   *
   4    4   * The contents of this file are subject to the terms of the
   5    5   * Common Development and Distribution License (the "License").
   6    6   * You may not use this file except in compliance with the License.
   7    7   *
   8    8   * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
   9    9   * or http://www.opensolaris.org/os/licensing.
  10   10   * See the License for the specific language governing permissions
  11   11   * and limitations under the License.
  12   12   *
  13   13   * When distributing Covered Code, include this CDDL HEADER in each
  14   14   * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
  15   15   * If applicable, add the following below this CDDL HEADER, with the
  16   16   * fields enclosed by brackets "[]" replaced with your own identifying
  17   17   * information: Portions Copyright [yyyy] [name of copyright owner]
  18   18   *

  19   19   * CDDL HEADER END
  20   20   */
  21   21  /*
  22   22   * Copyright 2011 Nexenta Systems, Inc.  All rights reserved.
  23   23   */
  24   24  /*
  25   25   * Copyright 2006 Sun Microsystems, Inc.  All rights reserved.
  26   26   * Use is subject to license terms.
  27   27   */
  28   28  
  29      -#pragma weak exp10 = __exp10
  30      -
  31   29  /* INDENT OFF */
  32   30  /*
  33   31   * exp10(x)
  34   32   * Code by K.C. Ng for SUN 4.0 libm.
  35   33   * Method :
  36   34   *      n = nint(x*(log10/log2));
  37   35   *      exp10(x) = 10**x = exp(x*ln(10)) = exp(n*ln2+(x*ln10-n*ln2))
  38   36   *               = 2**n*exp(ln10*(x-n*log2/log10)))
  39   37   *      If x is an integer < 23 then use repeat multiplication. For
  40   38   *      10**22 is the largest representable integer.

  41   39   */
  42   40  /* INDENT ON */
  43   41  
  44   42  #include "libm.h"
  45   43  
  46   44  static const double C[] = {
  47   45          3.3219280948736234787,  /* log(10)/log(2) */
  48   46          2.3025850929940456840,  /* log(10) */
  49   47          3.0102999565860955045E-1,       /* log(2)/log(10) high */
  50   48          5.3716447674669983622E-12,      /* log(2)/log(10) low */
  51   49          0.0,
  52   50          0.5,
  53   51          1.0,
  54   52          10.0,
  55   53          1.0e300,
  56   54          1.0e-300,
  57   55  };
  58   56  
  59   57  #define lg10    C[0]
  60   58  #define ln10    C[1]
  61   59  #define logt2hi C[2]
  62   60  #define logt2lo C[3]
  63   61  #define zero    C[4]
  64   62  #define half    C[5]
  65   63  #define one     C[6]
  66   64  #define ten     C[7]
  67   65  #define huge    C[8]
  68   66  #define tiny    C[9]
  69   67  
  70   68  double
  71   69  exp10(double x) {
  72   70          double  t, pt;
  73   71          int     ix, hx, k;
  74   72  
  75   73          ix = ((int *)&x)[HIWORD];
  76   74          hx = ix & ~0x80000000;
  77   75  
  78   76          if (hx >= 0x4074a000) { /* |x| >= 330 or x is nan */
  79   77                  if (hx >= 0x7ff00000) { /* x is inf or nan */
  80   78                          if (ix == 0xfff00000 && ((int *)&x)[LOWORD] == 0)
  81   79                                  return (zero);
  82   80                          return (x * x);
  83   81                  }
  84   82                  t = (ix < 0)? tiny : huge;
  85   83                  return (t * t);
  86   84          }
  87   85  
  88   86          if (hx < 0x3c000000)
  89   87                  return (one + x);
  90   88  
  91   89          k = (int)x;
  92   90          if (0 <= k && k < 23 && (double)k == x) {
  93   91                  /* x is a small positive integer */
  94   92                  t = one;
  95   93                  pt = ten;
  96   94                  if (k & 1)
  97   95                          t = ten;
  98   96                  k >>= 1;
  99   97                  while (k) {
 100   98                          pt *= pt;
 101   99                          if (k & 1)
 102  100                                  t *= pt;
 103  101                          k >>= 1;
 104  102                  }
 105  103                  return (t);
 106  104          }
 107  105          t = x * lg10;
 108  106          k = (int)((ix < 0)? t - half : t + half);
 109  107          return (scalbn(exp(ln10 * ((x - k * logt2hi) - k * logt2lo)), k));
 110  108  }

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