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121 lines
3.7 KiB
C
121 lines
3.7 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
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ln2 = 0.69314718055994530942,
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two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */
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Lg1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
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Lg2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
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Lg3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
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Lg4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
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Lg5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
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Lg6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
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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) { /* 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; 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)) 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) { /* |f| < 2**-20 */
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if(f==zero) 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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hfsq=0.5*f*f;
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return dk-((hfsq-(s*(hfsq+R)))-f)/ln2;
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} else {
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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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