Update.

1 files changed, 0 insertions, 452 deletions

diff --git a/sysdeps/libm-ieee754/e_sqrt.c b/sysdeps/libm-ieee754/e_sqrt.c
deleted file mode 100644
index 67da5455f9..0000000000
--- a/sysdeps/libm-ieee754/e_sqrt.c
+++ /dev/null
@@ -1,452 +0,0 @@
-/* @(#)e_sqrt.c 5.1 93/09/24 */
-/*
- * ====================================================
- * 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.
- * ====================================================
- */
-
-#if defined(LIBM_SCCS) && !defined(lint)
-static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
-#endif
-
-/* __ieee754_sqrt(x)
- * Return correctly rounded sqrt.
- *           ------------------------------------------
- *	     |  Use the hardware sqrt if you have one |
- *           ------------------------------------------
- * Method:
- *   Bit by bit method using integer arithmetic. (Slow, but portable)
- *   1. Normalization
- *	Scale x to y in [1,4) with even powers of 2:
- *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
- *		sqrt(x) = 2^k * sqrt(y)
- *   2. Bit by bit computation
- *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
- *	     i							 0
- *                                     i+1         2
- *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
- *	     i      i            i                 i
- *
- *	To compute q    from q , one checks whether
- *		    i+1       i
- *
- *			      -(i+1) 2
- *			(q + 2      ) <= y.			(2)
- *     			  i
- *							      -(i+1)
- *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
- *		 	       i+1   i             i+1   i
- *
- *	With some algebraic manipulation, it is not difficult to see
- *	that (2) is equivalent to
- *                             -(i+1)
- *			s  +  2       <= y			(3)
- *			 i                i
- *
- *	The advantage of (3) is that s  and y  can be computed by
- *				      i      i
- *	the following recurrence formula:
- *	    if (3) is false
- *
- *	    s     =  s  ,	y    = y   ;			(4)
- *	     i+1      i		 i+1    i
- *
- *	    otherwise,
- *                         -i                     -(i+1)
- *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
- *           i+1      i          i+1    i     i
- *
- *	One may easily use induction to prove (4) and (5).
- *	Note. Since the left hand side of (3) contain only i+2 bits,
- *	      it does not necessary to do a full (53-bit) comparison
- *	      in (3).
- *   3. Final rounding
- *	After generating the 53 bits result, we compute one more bit.
- *	Together with the remainder, we can decide whether the
- *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
- *	(it will never equal to 1/2ulp).
- *	The rounding mode can be detected by checking whether
- *	huge + tiny is equal to huge, and whether huge - tiny is
- *	equal to huge for some floating point number "huge" and "tiny".
- *
- * Special cases:
- *	sqrt(+-0) = +-0 	... exact
- *	sqrt(inf) = inf
- *	sqrt(-ve) = NaN		... with invalid signal
- *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
- *
- * Other methods : see the appended file at the end of the program below.
- *---------------
- */
-
-#include "math.h"
-#include "math_private.h"
-
-#ifdef __STDC__
-static	const double	one	= 1.0, tiny=1.0e-300;
-#else
-static	double	one	= 1.0, tiny=1.0e-300;
-#endif
-
-#ifdef __STDC__
-	double __ieee754_sqrt(double x)
-#else
-	double __ieee754_sqrt(x)
-	double x;
-#endif
-{
-	double z;
-	int32_t sign = (int)0x80000000;
-	int32_t ix0,s0,q,m,t,i;
-	u_int32_t r,t1,s1,ix1,q1;
-
-	EXTRACT_WORDS(ix0,ix1,x);
-
-    /* take care of Inf and NaN */
-	if((ix0&0x7ff00000)==0x7ff00000) {
-	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
-					   sqrt(-inf)=sNaN */
-	}
-    /* take care of zero */
-	if(ix0<=0) {
-	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
-	    else if(ix0<0)
-		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
-	}
-    /* normalize x */
-	m = (ix0>>20);
-	if(m==0) {				/* subnormal x */
-	    while(ix0==0) {
-		m -= 21;
-		ix0 |= (ix1>>11); ix1 <<= 21;
-	    }
-	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
-	    m -= i-1;
-	    ix0 |= (ix1>>(32-i));
-	    ix1 <<= i;
-	}
-	m -= 1023;	/* unbias exponent */
-	ix0 = (ix0&0x000fffff)|0x00100000;
-	if(m&1){	/* odd m, double x to make it even */
-	    ix0 += ix0 + ((ix1&sign)>>31);
-	    ix1 += ix1;
-	}
-	m >>= 1;	/* m = [m/2] */
-
-    /* generate sqrt(x) bit by bit */
-	ix0 += ix0 + ((ix1&sign)>>31);
-	ix1 += ix1;
-	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
-	r = 0x00200000;		/* r = moving bit from right to left */
-
-	while(r!=0) {
-	    t = s0+r;
-	    if(t<=ix0) {
-		s0   = t+r;
-		ix0 -= t;
-		q   += r;
-	    }
-	    ix0 += ix0 + ((ix1&sign)>>31);
-	    ix1 += ix1;
-	    r>>=1;
-	}
-
-	r = sign;
-	while(r!=0) {
-	    t1 = s1+r;
-	    t  = s0;
-	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
-		s1  = t1+r;
-		if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
-		ix0 -= t;
-		if (ix1 < t1) ix0 -= 1;
-		ix1 -= t1;
-		q1  += r;
-	    }
-	    ix0 += ix0 + ((ix1&sign)>>31);
-	    ix1 += ix1;
-	    r>>=1;
-	}
-
-    /* use floating add to find out rounding direction */
-	if((ix0|ix1)!=0) {
-	    z = one-tiny; /* trigger inexact flag */
-	    if (z>=one) {
-	        z = one+tiny;
-	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
-		else if (z>one) {
-		    if (q1==(u_int32_t)0xfffffffe) q+=1;
-		    q1+=2;
-		} else
-	            q1 += (q1&1);
-	    }
-	}
-	ix0 = (q>>1)+0x3fe00000;
-	ix1 =  q1>>1;
-	if ((q&1)==1) ix1 |= sign;
-	ix0 += (m <<20);
-	INSERT_WORDS(z,ix0,ix1);
-	return z;
-}
-
-/*
-Other methods  (use floating-point arithmetic)
--------------
-(This is a copy of a drafted paper by Prof W. Kahan
-and K.C. Ng, written in May, 1986)
-
-	Two algorithms are given here to implement sqrt(x)
-	(IEEE double precision arithmetic) in software.
-	Both supply sqrt(x) correctly rounded. The first algorithm (in
-	Section A) uses newton iterations and involves four divisions.
-	The second one uses reciproot iterations to avoid division, but
-	requires more multiplications. Both algorithms need the ability
-	to chop results of arithmetic operations instead of round them,
-	and the INEXACT flag to indicate when an arithmetic operation
-	is executed exactly with no roundoff error, all part of the
-	standard (IEEE 754-1985). The ability to perform shift, add,
-	subtract and logical AND operations upon 32-bit words is needed
-	too, though not part of the standard.
-
-A.  sqrt(x) by Newton Iteration
-
-   (1)	Initial approximation
-
-	Let x0 and x1 be the leading and the trailing 32-bit words of
-	a floating point number x (in IEEE double format) respectively
-
-	    1    11		     52				  ...widths
-	   ------------------------------------------------------
-	x: |s|	  e     |	      f				|
-	   ------------------------------------------------------
-	      msb    lsb  msb				      lsb ...order
-
-
-	     ------------------------  	     ------------------------
-	x0:  |s|   e    |    f1     |	 x1: |          f2           |
-	     ------------------------  	     ------------------------
-
-	By performing shifts and subtracts on x0 and x1 (both regarded
-	as integers), we obtain an 8-bit approximation of sqrt(x) as
-	follows.
-
-		k  := (x0>>1) + 0x1ff80000;
-		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
-	Here k is a 32-bit integer and T1[] is an integer array containing
-	correction terms. Now magically the floating value of y (y's
-	leading 32-bit word is y0, the value of its trailing word is 0)
-	approximates sqrt(x) to almost 8-bit.
-
-	Value of T1:
-	static int T1[32]= {
-	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
-	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
-	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
-	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
-
-    (2)	Iterative refinement
-
-	Apply Heron's rule three times to y, we have y approximates
-	sqrt(x) to within 1 ulp (Unit in the Last Place):
-
-		y := (y+x/y)/2		... almost 17 sig. bits
-		y := (y+x/y)/2		... almost 35 sig. bits
-		y := y-(y-x/y)/2	... within 1 ulp
-
-
-	Remark 1.
-	    Another way to improve y to within 1 ulp is:
-
-		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
-		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
-
-				2
-			    (x-y )*y
-		y := y + 2* ----------	...within 1 ulp
-			       2
-			     3y  + x
-
-
-	This formula has one division fewer than the one above; however,
-	it requires more multiplications and additions. Also x must be
-	scaled in advance to avoid spurious overflow in evaluating the
-	expression 3y*y+x. Hence it is not recommended uless division
-	is slow. If division is very slow, then one should use the
-	reciproot algorithm given in section B.
-
-    (3) Final adjustment
-
-	By twiddling y's last bit it is possible to force y to be
-	correctly rounded according to the prevailing rounding mode
-	as follows. Let r and i be copies of the rounding mode and
-	inexact flag before entering the square root program. Also we
-	use the expression y+-ulp for the next representable floating
-	numbers (up and down) of y. Note that y+-ulp = either fixed
-	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
-	mode.
-
-		I := FALSE;	... reset INEXACT flag I
-		R := RZ;	... set rounding mode to round-toward-zero
-		z := x/y;	... chopped quotient, possibly inexact
-		If(not I) then {	... if the quotient is exact
-		    if(z=y) {
-		        I := i;	 ... restore inexact flag
-		        R := r;  ... restore rounded mode
-		        return sqrt(x):=y.
-		    } else {
-			z := z - ulp;	... special rounding
-		    }
-		}
-		i := TRUE;		... sqrt(x) is inexact
-		If (r=RN) then z=z+ulp	... rounded-to-nearest
-		If (r=RP) then {	... round-toward-+inf
-		    y = y+ulp; z=z+ulp;
-		}
-		y := y+z;		... chopped sum
-		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
-	        I := i;	 		... restore inexact flag
-	        R := r;  		... restore rounded mode
-	        return sqrt(x):=y.
-
-    (4)	Special cases
-
-	Square root of +inf, +-0, or NaN is itself;
-	Square root of a negative number is NaN with invalid signal.
-
-
-B.  sqrt(x) by Reciproot Iteration
-
-   (1)	Initial approximation
-
-	Let x0 and x1 be the leading and the trailing 32-bit words of
-	a floating point number x (in IEEE double format) respectively
-	(see section A). By performing shifs and subtracts on x0 and y0,
-	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
-
-	    k := 0x5fe80000 - (x0>>1);
-	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
-
-	Here k is a 32-bit integer and T2[] is an integer array
-	containing correction terms. Now magically the floating
-	value of y (y's leading 32-bit word is y0, the value of
-	its trailing word y1 is set to zero) approximates 1/sqrt(x)
-	to almost 7.8-bit.
-
-	Value of T2:
-	static int T2[64]= {
-	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
-	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
-	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
-	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
-	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
-	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
-	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
-	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
-
-    (2)	Iterative refinement
-
-	Apply Reciproot iteration three times to y and multiply the
-	result by x to get an approximation z that matches sqrt(x)
-	to about 1 ulp. To be exact, we will have
-		-1ulp < sqrt(x)-z<1.0625ulp.
-
-	... set rounding mode to Round-to-nearest
-	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
-	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
-	... special arrangement for better accuracy
-	   z := x*y			... 29 bits to sqrt(x), with z*y<1
-	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
-
-	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
-	(a) the term z*y in the final iteration is always less than 1;
-	(b) the error in the final result is biased upward so that
-		-1 ulp < sqrt(x) - z < 1.0625 ulp
-	    instead of |sqrt(x)-z|<1.03125ulp.
-
-    (3)	Final adjustment
-
-	By twiddling y's last bit it is possible to force y to be
-	correctly rounded according to the prevailing rounding mode
-	as follows. Let r and i be copies of the rounding mode and
-	inexact flag before entering the square root program. Also we
-	use the expression y+-ulp for the next representable floating
-	numbers (up and down) of y. Note that y+-ulp = either fixed
-	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
-	mode.
-
-	R := RZ;		... set rounding mode to round-toward-zero
-	switch(r) {
-	    case RN:		... round-to-nearest
-	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
-	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
-	       break;
-	    case RZ:case RM:	... round-to-zero or round-to--inf
-	       R:=RP;		... reset rounding mod to round-to-+inf
-	       if(x<z*z ... rounded up) z = z - ulp; else
-	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
-	       break;
-	    case RP:		... round-to-+inf
-	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
-	       if(x>z*z ...chopped) z = z+ulp;
-	       break;
-	}
-
-	Remark 3. The above comparisons can be done in fixed point. For
-	example, to compare x and w=z*z chopped, it suffices to compare
-	x1 and w1 (the trailing parts of x and w), regarding them as
-	two's complement integers.
-
-	...Is z an exact square root?
-	To determine whether z is an exact square root of x, let z1 be the
-	trailing part of z, and also let x0 and x1 be the leading and
-	trailing parts of x.
-
-	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
-	    I := 1;		... Raise Inexact flag: z is not exact
-	else {
-	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
-	    k := z1 >> 26;		... get z's 25-th and 26-th
-					    fraction bits
-	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
-	}
-	R:= r		... restore rounded mode
-	return sqrt(x):=z.
-
-	If multiplication is cheaper then the foregoing red tape, the
-	Inexact flag can be evaluated by
-
-	    I := i;
-	    I := (z*z!=x) or I.
-
-	Note that z*z can overwrite I; this value must be sensed if it is
-	True.
-
-	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
-	zero.
-
-		    --------------------
-		z1: |        f2        |
-		    --------------------
-		bit 31		   bit 0
-
-	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
-	or even of logb(x) have the following relations:
-
-	-------------------------------------------------
-	bit 27,26 of z1		bit 1,0 of x1	logb(x)
-	-------------------------------------------------
-	00			00		odd and even
-	01			01		even
-	10			10		odd
-	10			00		even
-	11			01		even
-	-------------------------------------------------
-
-    (4)	Special cases (see (4) of Section A).
-
- */