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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185
|
/* 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;
ix0 += m << 20;
INSERT_WORDS(z, ix0, ix1);
return z;
}
|