185 lines
5.3 KiB
C
185 lines
5.3 KiB
C
/* origin: FreeBSD /usr/src/lib/msun/src/e_sqrt.c */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunSoft, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* sqrt(x)
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* Return correctly rounded sqrt.
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* ------------------------------------------
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* | Use the hardware sqrt if you have one |
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* ------------------------------------------
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* Method:
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* Bit by bit method using integer arithmetic. (Slow, but portable)
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* 1. Normalization
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* Scale x to y in [1,4) with even powers of 2:
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* find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
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* sqrt(x) = 2^k * sqrt(y)
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* 2. Bit by bit computation
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* Let q = sqrt(y) truncated to i bit after binary point (q = 1),
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* i 0
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* i+1 2
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* s = 2*q , and y = 2 * ( y - q ). (1)
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* i i i i
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*
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* To compute q from q , one checks whether
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* i+1 i
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*
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* -(i+1) 2
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* (q + 2 ) <= y. (2)
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* i
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* -(i+1)
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* If (2) is false, then q = q ; otherwise q = q + 2 .
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* i+1 i i+1 i
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*
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* With some algebric manipulation, it is not difficult to see
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* that (2) is equivalent to
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* -(i+1)
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* s + 2 <= y (3)
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* i i
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*
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* The advantage of (3) is that s and y can be computed by
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* i i
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* the following recurrence formula:
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* if (3) is false
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*
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* s = s , y = y ; (4)
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* i+1 i i+1 i
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*
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* otherwise,
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* -i -(i+1)
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* s = s + 2 , y = y - s - 2 (5)
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* i+1 i i+1 i i
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*
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* One may easily use induction to prove (4) and (5).
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* Note. Since the left hand side of (3) contain only i+2 bits,
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* it does not necessary to do a full (53-bit) comparison
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* in (3).
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* 3. Final rounding
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* After generating the 53 bits result, we compute one more bit.
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* Together with the remainder, we can decide whether the
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* result is exact, bigger than 1/2ulp, or less than 1/2ulp
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* (it will never equal to 1/2ulp).
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* The rounding mode can be detected by checking whether
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* huge + tiny is equal to huge, and whether huge - tiny is
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* equal to huge for some floating point number "huge" and "tiny".
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*
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* Special cases:
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* sqrt(+-0) = +-0 ... exact
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* sqrt(inf) = inf
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* sqrt(-ve) = NaN ... with invalid signal
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* sqrt(NaN) = NaN ... with invalid signal for signaling NaN
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*/
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#include "libc/math/libm.h"
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static const double tiny = 1.0e-300;
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double sqrt(double x)
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{
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double z;
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int32_t sign = (int)0x80000000;
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int32_t ix0,s0,q,m,t,i;
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uint32_t r,t1,s1,ix1,q1;
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EXTRACT_WORDS(ix0, ix1, x);
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/* take care of Inf and NaN */
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if ((ix0&0x7ff00000) == 0x7ff00000) {
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return x*x + x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf, sqrt(-inf)=sNaN */
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}
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/* take care of zero */
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if (ix0 <= 0) {
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if (((ix0&~sign)|ix1) == 0)
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return x; /* sqrt(+-0) = +-0 */
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if (ix0 < 0)
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return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
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}
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/* normalize x */
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m = ix0>>20;
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if (m == 0) { /* subnormal x */
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while (ix0 == 0) {
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m -= 21;
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ix0 |= (ix1>>11);
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ix1 <<= 21;
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}
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for (i=0; (ix0&0x00100000) == 0; i++)
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ix0<<=1;
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m -= i - 1;
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ix0 |= ix1>>(32-i);
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ix1 <<= i;
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}
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m -= 1023; /* unbias exponent */
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ix0 = (ix0&0x000fffff)|0x00100000;
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if (m & 1) { /* odd m, double x to make it even */
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ix0 += ix0 + ((ix1&sign)>>31);
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ix1 += ix1;
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}
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m >>= 1; /* m = [m/2] */
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/* generate sqrt(x) bit by bit */
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ix0 += ix0 + ((ix1&sign)>>31);
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ix1 += ix1;
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q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
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r = 0x00200000; /* r = moving bit from right to left */
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while (r != 0) {
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t = s0 + r;
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if (t <= ix0) {
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s0 = t + r;
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ix0 -= t;
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q += r;
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}
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ix0 += ix0 + ((ix1&sign)>>31);
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ix1 += ix1;
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r >>= 1;
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}
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r = sign;
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while (r != 0) {
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t1 = s1 + r;
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t = s0;
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if (t < ix0 || (t == ix0 && t1 <= ix1)) {
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s1 = t1 + r;
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if ((t1&sign) == sign && (s1&sign) == 0)
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s0++;
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ix0 -= t;
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if (ix1 < t1)
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ix0--;
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ix1 -= t1;
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q1 += r;
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}
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ix0 += ix0 + ((ix1&sign)>>31);
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ix1 += ix1;
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r >>= 1;
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}
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/* use floating add to find out rounding direction */
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if ((ix0|ix1) != 0) {
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z = 1.0 - tiny; /* raise inexact flag */
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if (z >= 1.0) {
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z = 1.0 + tiny;
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if (q1 == (uint32_t)0xffffffff) {
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q1 = 0;
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q++;
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} else if (z > 1.0) {
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if (q1 == (uint32_t)0xfffffffe)
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q++;
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q1 += 2;
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} else
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q1 += q1 & 1;
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}
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}
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ix0 = (q>>1) + 0x3fe00000;
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ix1 = q1>>1;
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if (q&1)
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ix1 |= sign;
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INSERT_WORDS(z, ix0 + ((uint32_t)m << 20), ix1);
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return z;
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}
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