1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
| | /* origin: FreeBSD /usr/src/lib/msun/src/e_sqrt.c */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
/* 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 algebric 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
*/
#include "libm.h"
static const double tiny = 1.0e-300;
double sqrt(double x)
{
double z;
int32_t sign = (int)0x80000000;
int32_t ix0,s0,q,m,t,i;
uint32_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 */
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++;
ix0 -= t;
if (ix1 < t1)
ix0--;
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 = 1.0 - tiny; /* raise inexact flag */
if (z >= 1.0) {
z = 1.0 + tiny;
if (q1 == (uint32_t)0xffffffff) {
q1 = 0;
q++;
} else if (z > 1.0) {
if (q1 == (uint32_t)0xfffffffe)
q++;
q1 += 2;
} else
q1 += q1 & 1;
}
}
ix0 = (q>>1) + 0x3fe00000;
ix1 = q1>>1;
if (q&1)
ix1 |= sign;
INSERT_WORDS(z, ix0 + ((uint32_t)m << 20), ix1);
return z;
}
|