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129 lines
4.1 KiB
C
129 lines
4.1 KiB
C
/* Adapted for log2 by Ulrich Drepper <drepper@cygnus.com>. */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* __ieee754_log2(x)
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* Return the logarithm to base 2 of x
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*
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* Method :
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* 1. Argument Reduction: find k and f such that
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* x = 2^k * (1+f),
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* where sqrt(2)/2 < 1+f < sqrt(2) .
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*
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* 2. Approximation of log(1+f).
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* Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
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* = 2s + 2/3 s**3 + 2/5 s**5 + .....,
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* = 2s + s*R
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* We use a special Reme algorithm on [0,0.1716] to generate
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* a polynomial of degree 14 to approximate R The maximum error
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* of this polynomial approximation is bounded by 2**-58.45. In
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* other words,
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* 2 4 6 8 10 12 14
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* R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s +Lg6*s +Lg7*s
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* (the values of Lg1 to Lg7 are listed in the program)
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* and
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* | 2 14 | -58.45
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* | Lg1*s +...+Lg7*s - R(z) | <= 2
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* | |
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* Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
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* In order to guarantee error in log below 1ulp, we compute log
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* by
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* log(1+f) = f - s*(f - R) (if f is not too large)
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* log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
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*
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* 3. Finally, log(x) = k + log(1+f).
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* = k+(f-(hfsq-(s*(hfsq+R))))
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*
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* Special cases:
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* log2(x) is NaN with signal if x < 0 (including -INF) ;
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* log2(+INF) is +INF; log(0) is -INF with signal;
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* log2(NaN) is that NaN with no signal.
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*
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* Constants:
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* The hexadecimal values are the intended ones for the following
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* constants. The decimal values may be used, provided that the
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* compiler will convert from decimal to binary accurately enough
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* to produce the hexadecimal values shown.
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*/
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#include <math.h>
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#include <math_private.h>
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static const double ln2 = 0.69314718055994530942;
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static const double two54 = 1.80143985094819840000e+16; /* 43500000 00000000 */
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static const double Lg1 = 6.666666666666735130e-01; /* 3FE55555 55555593 */
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static const double Lg2 = 3.999999999940941908e-01; /* 3FD99999 9997FA04 */
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static const double Lg3 = 2.857142874366239149e-01; /* 3FD24924 94229359 */
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static const double Lg4 = 2.222219843214978396e-01; /* 3FCC71C5 1D8E78AF */
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static const double Lg5 = 1.818357216161805012e-01; /* 3FC74664 96CB03DE */
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static const double Lg6 = 1.531383769920937332e-01; /* 3FC39A09 D078C69F */
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static const double Lg7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
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static const double zero = 0.0;
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double
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__ieee754_log2 (double x)
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{
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double hfsq, f, s, z, R, w, t1, t2, dk;
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int32_t k, hx, i, j;
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u_int32_t lx;
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EXTRACT_WORDS (hx, lx, x);
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k = 0;
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if (hx < 0x00100000)
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{ /* x < 2**-1022 */
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if (__builtin_expect (((hx & 0x7fffffff) | lx) == 0, 0))
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return -two54 / (x - x); /* log(+-0)=-inf */
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if (__builtin_expect (hx < 0, 0))
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return (x - x) / (x - x); /* log(-#) = NaN */
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k -= 54;
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x *= two54; /* subnormal number, scale up x */
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GET_HIGH_WORD (hx, x);
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}
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if (__builtin_expect (hx >= 0x7ff00000, 0))
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return x + x;
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k += (hx >> 20) - 1023;
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hx &= 0x000fffff;
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i = (hx + 0x95f64) & 0x100000;
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SET_HIGH_WORD (x, hx | (i ^ 0x3ff00000)); /* normalize x or x/2 */
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k += (i >> 20);
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dk = (double) k;
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f = x - 1.0;
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if ((0x000fffff & (2 + hx)) < 3)
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{ /* |f| < 2**-20 */
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if (f == zero)
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return dk;
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R = f * f * (0.5 - 0.33333333333333333 * f);
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return dk - (R - f) / ln2;
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}
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s = f / (2.0 + f);
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z = s * s;
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i = hx - 0x6147a;
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w = z * z;
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j = 0x6b851 - hx;
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t1 = w * (Lg2 + w * (Lg4 + w * Lg6));
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t2 = z * (Lg1 + w * (Lg3 + w * (Lg5 + w * Lg7)));
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i |= j;
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R = t2 + t1;
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if (i > 0)
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{
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hfsq = 0.5 * f * f;
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return dk - ((hfsq - (s * (hfsq + R))) - f) / ln2;
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}
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else
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{
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return dk - ((s * (f - R)) - f) / ln2;
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}
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}
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strong_alias (__ieee754_log2, __log2_finite)
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