source: asp3_tinet_ecnl_rx/trunk/musl-1.1.18/src/math/sqrt.c@ 337

Last change on this file since 337 was 337, checked in by coas-nagasima, 6 years ago

ASP3版ECNLを追加
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1/* origin: FreeBSD /usr/src/lib/msun/src/e_sqrt.c */
2/*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
9 * is preserved.
10 * ====================================================
11 */
12/* sqrt(x)
13 * Return correctly rounded sqrt.
14 * ------------------------------------------
15 * | Use the hardware sqrt if you have one |
16 * ------------------------------------------
17 * Method:
18 * Bit by bit method using integer arithmetic. (Slow, but portable)
19 * 1. Normalization
20 * Scale x to y in [1,4) with even powers of 2:
21 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
22 * sqrt(x) = 2^k * sqrt(y)
23 * 2. Bit by bit computation
24 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
25 * i 0
26 * i+1 2
27 * s = 2*q , and y = 2 * ( y - q ). (1)
28 * i i i i
29 *
30 * To compute q from q , one checks whether
31 * i+1 i
32 *
33 * -(i+1) 2
34 * (q + 2 ) <= y. (2)
35 * i
36 * -(i+1)
37 * If (2) is false, then q = q ; otherwise q = q + 2 .
38 * i+1 i i+1 i
39 *
40 * With some algebric manipulation, it is not difficult to see
41 * that (2) is equivalent to
42 * -(i+1)
43 * s + 2 <= y (3)
44 * i i
45 *
46 * The advantage of (3) is that s and y can be computed by
47 * i i
48 * the following recurrence formula:
49 * if (3) is false
50 *
51 * s = s , y = y ; (4)
52 * i+1 i i+1 i
53 *
54 * otherwise,
55 * -i -(i+1)
56 * s = s + 2 , y = y - s - 2 (5)
57 * i+1 i i+1 i i
58 *
59 * One may easily use induction to prove (4) and (5).
60 * Note. Since the left hand side of (3) contain only i+2 bits,
61 * it does not necessary to do a full (53-bit) comparison
62 * in (3).
63 * 3. Final rounding
64 * After generating the 53 bits result, we compute one more bit.
65 * Together with the remainder, we can decide whether the
66 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
67 * (it will never equal to 1/2ulp).
68 * The rounding mode can be detected by checking whether
69 * huge + tiny is equal to huge, and whether huge - tiny is
70 * equal to huge for some floating point number "huge" and "tiny".
71 *
72 * Special cases:
73 * sqrt(+-0) = +-0 ... exact
74 * sqrt(inf) = inf
75 * sqrt(-ve) = NaN ... with invalid signal
76 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
77 */
78
79#include "libm.h"
80
81static const double tiny = 1.0e-300;
82
83double sqrt(double x)
84{
85 double z;
86 int32_t sign = (int)0x80000000;
87 int32_t ix0,s0,q,m,t,i;
88 uint32_t r,t1,s1,ix1,q1;
89
90 EXTRACT_WORDS(ix0, ix1, x);
91
92 /* take care of Inf and NaN */
93 if ((ix0&0x7ff00000) == 0x7ff00000) {
94 return x*x + x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf, sqrt(-inf)=sNaN */
95 }
96 /* take care of zero */
97 if (ix0 <= 0) {
98 if (((ix0&~sign)|ix1) == 0)
99 return x; /* sqrt(+-0) = +-0 */
100 if (ix0 < 0)
101 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
102 }
103 /* normalize x */
104 m = ix0>>20;
105 if (m == 0) { /* subnormal x */
106 while (ix0 == 0) {
107 m -= 21;
108 ix0 |= (ix1>>11);
109 ix1 <<= 21;
110 }
111 for (i=0; (ix0&0x00100000) == 0; i++)
112 ix0<<=1;
113 m -= i - 1;
114 ix0 |= ix1>>(32-i);
115 ix1 <<= i;
116 }
117 m -= 1023; /* unbias exponent */
118 ix0 = (ix0&0x000fffff)|0x00100000;
119 if (m & 1) { /* odd m, double x to make it even */
120 ix0 += ix0 + ((ix1&sign)>>31);
121 ix1 += ix1;
122 }
123 m >>= 1; /* m = [m/2] */
124
125 /* generate sqrt(x) bit by bit */
126 ix0 += ix0 + ((ix1&sign)>>31);
127 ix1 += ix1;
128 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
129 r = 0x00200000; /* r = moving bit from right to left */
130
131 while (r != 0) {
132 t = s0 + r;
133 if (t <= ix0) {
134 s0 = t + r;
135 ix0 -= t;
136 q += r;
137 }
138 ix0 += ix0 + ((ix1&sign)>>31);
139 ix1 += ix1;
140 r >>= 1;
141 }
142
143 r = sign;
144 while (r != 0) {
145 t1 = s1 + r;
146 t = s0;
147 if (t < ix0 || (t == ix0 && t1 <= ix1)) {
148 s1 = t1 + r;
149 if ((t1&sign) == sign && (s1&sign) == 0)
150 s0++;
151 ix0 -= t;
152 if (ix1 < t1)
153 ix0--;
154 ix1 -= t1;
155 q1 += r;
156 }
157 ix0 += ix0 + ((ix1&sign)>>31);
158 ix1 += ix1;
159 r >>= 1;
160 }
161
162 /* use floating add to find out rounding direction */
163 if ((ix0|ix1) != 0) {
164 z = 1.0 - tiny; /* raise inexact flag */
165 if (z >= 1.0) {
166 z = 1.0 + tiny;
167 if (q1 == (uint32_t)0xffffffff) {
168 q1 = 0;
169 q++;
170 } else if (z > 1.0) {
171 if (q1 == (uint32_t)0xfffffffe)
172 q++;
173 q1 += 2;
174 } else
175 q1 += q1 & 1;
176 }
177 }
178 ix0 = (q>>1) + 0x3fe00000;
179 ix1 = q1>>1;
180 if (q&1)
181 ix1 |= sign;
182 ix0 += m << 20;
183 INSERT_WORDS(z, ix0, ix1);
184 return z;
185}
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