uclibc-ng/libm/e_log2.c

/* Adapted for log2 by Ulrich Drepper <drepper@cygnus.com>.  */
/*
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * is preserved.
 * ====================================================
 */

/* __ieee754_log2(x)
 * Return the logarithm to base 2 of x
 *
 * Method :
 *   1. Argument Reduction: find k and f such that
 *			x = 2^k * (1+f),
 *	   where  sqrt(2)/2 < 1+f < sqrt(2) .
 *
 *   2. Approximation of log(1+f).
 *	Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
 *		 = 2s + 2/3 s**3 + 2/5 s**5 + .....,
 *	     	 = 2s + s*R
 *      We use a special Reme algorithm on [0,0.1716] to generate
 * 	a polynomial of degree 14 to approximate R The maximum error
 *	of this polynomial approximation is bounded by 2**-58.45. In
 *	other words,
 *		        2      4      6      8      10      12      14
 *	    R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s  +Lg6*s  +Lg7*s
 *  	(the values of Lg1 to Lg7 are listed in the program)
 *	and
 *	    |      2          14          |     -58.45
 *	    | Lg1*s +...+Lg7*s    -  R(z) | <= 2
 *	    |                             |
 *	Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
 *	In order to guarantee error in log below 1ulp, we compute log
 *	by
 *		log(1+f) = f - s*(f - R)	(if f is not too large)
 *		log(1+f) = f - (hfsq - s*(hfsq+R)).	(better accuracy)
 *
 *	3. Finally,  log(x) = k + log(1+f).
 *			    = k+(f-(hfsq-(s*(hfsq+R))))
 *
 * Special cases:
 *	log2(x) is NaN with signal if x < 0 (including -INF) ;
 *	log2(+INF) is +INF; log(0) is -INF with signal;
 *	log2(NaN) is that NaN with no signal.
 *
 * Constants:
 * The hexadecimal values are the intended ones for the following
 * constants. The decimal values may be used, provided that the
 * compiler will convert from decimal to binary accurately enough
 * to produce the hexadecimal values shown.
 */

#include "math.h"
#include "math_private.h"

static const double
ln2 = 0.69314718055994530942,
two54   =  1.80143985094819840000e+16,  /* 43500000 00000000 */
Lg1 = 6.666666666666735130e-01,  /* 3FE55555 55555593 */
Lg2 = 3.999999999940941908e-01,  /* 3FD99999 9997FA04 */
Lg3 = 2.857142874366239149e-01,  /* 3FD24924 94229359 */
Lg4 = 2.222219843214978396e-01,  /* 3FCC71C5 1D8E78AF */
Lg5 = 1.818357216161805012e-01,  /* 3FC74664 96CB03DE */
Lg6 = 1.531383769920937332e-01,  /* 3FC39A09 D078C69F */
Lg7 = 1.479819860511658591e-01;  /* 3FC2F112 DF3E5244 */

static const double zero   =  0.0;

double __ieee754_log2(double x)
{
	double hfsq,f,s,z,R,w,t1,t2,dk;
	int32_t k,hx,i,j;
	u_int32_t lx;

	EXTRACT_WORDS(hx,lx,x);

	k=0;
	if (hx < 0x00100000) {			/* x < 2**-1022  */
	    if (((hx&0x7fffffff)|lx)==0)
		return -two54/(x-x);		/* log(+-0)=-inf */
	    if (hx<0) return (x-x)/(x-x);	/* log(-#) = NaN */
	    k -= 54; x *= two54; /* subnormal number, scale up x */
	    GET_HIGH_WORD(hx,x);
	}
	if (hx >= 0x7ff00000) return x+x;
	k += (hx>>20)-1023;
	hx &= 0x000fffff;
	i = (hx+0x95f64)&0x100000;
	SET_HIGH_WORD(x,hx|(i^0x3ff00000));	/* normalize x or x/2 */
	k += (i>>20);
	dk = (double) k;
	f = x-1.0;
	if((0x000fffff&(2+hx))<3) {	/* |f| < 2**-20 */
	    if(f==zero) return dk;
	    R = f*f*(0.5-0.33333333333333333*f);
	    return dk-(R-f)/ln2;
	}
	s = f/(2.0+f);
	z = s*s;
	i = hx-0x6147a;
	w = z*z;
	j = 0x6b851-hx;
	t1= w*(Lg2+w*(Lg4+w*Lg6));
	t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
	i |= j;
	R = t2+t1;
	if(i>0) {
	    hfsq=0.5*f*f;
	    return dk-((hfsq-(s*(hfsq+R)))-f)/ln2;
	} else {
	    return dk-((s*(f-R))-f)/ln2;
	}
}