1 /** @file
2 Compute the base 10 logrithm of x.
3
4 Copyright (c) 2010 - 2011, Intel Corporation. All rights reserved.<BR>
5 This program and the accompanying materials are licensed and made available under
6 the terms and conditions of the BSD License that accompanies this distribution.
7 The full text of the license may be found at
8 http://opensource.org/licenses/bsd-license.
9
10 THE PROGRAM IS DISTRIBUTED UNDER THE BSD LICENSE ON AN "AS IS" BASIS,
11 WITHOUT WARRANTIES OR REPRESENTATIONS OF ANY KIND, EITHER EXPRESS OR IMPLIED.
12
13 * ====================================================
14 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
15 *
16 * Developed at SunPro, a Sun Microsystems, Inc. business.
17 * Permission to use, copy, modify, and distribute this
18 * software is freely granted, provided that this notice
19 * is preserved.
20 * ====================================================
21
22 e_log10.c 5.1 93/09/24
23 NetBSD: e_log10.c,v 1.12 2002/05/26 22:01:51 wiz Exp
24 **/
25 #include <LibConfig.h>
26 #include <sys/EfiCdefs.h>
27
28 /* __ieee754_log10(x)
29 * Return the base 10 logarithm of x
30 *
31 * Method :
32 * Let log10_2hi = leading 40 bits of log10(2) and
33 * log10_2lo = log10(2) - log10_2hi,
34 * ivln10 = 1/log(10) rounded.
35 * Then
36 * n = ilogb(x),
37 * if(n<0) n = n+1;
38 * x = scalbn(x,-n);
39 * log10(x) := n*log10_2hi + (n*log10_2lo + ivln10*log(x))
40 *
41 * Note 1:
42 * To guarantee log10(10**n)=n, where 10**n is normal, the rounding
43 * mode must set to Round-to-Nearest.
44 * Note 2:
45 * [1/log(10)] rounded to 53 bits has error .198 ulps;
46 * log10 is monotonic at all binary break points.
47 *
48 * Special cases:
49 * log10(x) is NaN with signal if x < 0;
50 * log10(+INF) is +INF with no signal; log10(0) is -INF with signal;
51 * log10(NaN) is that NaN with no signal;
52 * log10(10**N) = N for N=0,1,...,22.
53 *
54 * Constants:
55 * The hexadecimal values are the intended ones for the following constants.
56 * The decimal values may be used, provided that the compiler will convert
57 * from decimal to binary accurately enough to produce the hexadecimal values
58 * shown.
59 */
60
61 #include "math.h"
62 #include "math_private.h"
63 #include <errno.h>
64
65 #if defined(_MSC_VER) /* Handle Microsoft VC++ compiler specifics. */
66 // potential divide by 0 -- near line 80, (x-x)/zero is on purpose
67 #pragma warning ( disable : 4723 )
68 #endif
69
70 static const double
71 two54 = 1.80143985094819840000e+16, /* 0x43500000, 0x00000000 */
72 ivln10 = 4.34294481903251816668e-01, /* 0x3FDBCB7B, 0x1526E50E */
73 log10_2hi = 3.01029995663611771306e-01, /* 0x3FD34413, 0x509F6000 */
74 log10_2lo = 3.69423907715893078616e-13; /* 0x3D59FEF3, 0x11F12B36 */
75
76 static const double zero = 0.0;
77
78 double
__ieee754_log10(double x)79 __ieee754_log10(double x)
80 {
81 double y,z;
82 int32_t i,k,hx;
83 u_int32_t lx;
84
85 EXTRACT_WORDS(hx,lx,x);
86
87 k=0;
88 if (hx < 0x00100000) { /* x < 2**-1022 */
89 if (((hx&0x7fffffff)|lx)==0)
90 return -two54/zero; /* log(+-0)=-inf */
91 if (hx<0) {
92 errno = EDOM;
93 return (x-x)/zero; /* log(-#) = NaN */
94 }
95 k -= 54; x *= two54; /* subnormal number, scale up x */
96 GET_HIGH_WORD(hx,x);
97 }
98 if (hx >= 0x7ff00000) return x+x;
99 k += (hx>>20)-1023;
100 i = ((u_int32_t)k&0x80000000)>>31;
101 hx = (hx&0x000fffff)|((0x3ff-i)<<20);
102 y = (double)(k+i);
103 SET_HIGH_WORD(x,hx);
104 z = y*log10_2lo + ivln10*__ieee754_log(x);
105 return z+y*log10_2hi;
106 }
107