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 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
