source: soft/giet_vm/giet_libs/math/sqrt.c

#include "../math.h"
#include "math_private.h"

static const double one = 1.0, tiny = 1.0e-300;

/* this function was taken from the uCLibc library
 * Return correctly rounded sqrt.
 * 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
 *
 * Other methods : see the appended file at the end of the program below.
 *---------------
 */
static double __ieee754_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 */
        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==(uint32_t)0xffffffff) { q1=0; q += 1;}
            else if (z>one) {
                if (q1==(uint32_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;
}

double sqrt(double x)
{
#if GIET_USE_HARD_FLOAT
    double z;
    __asm__ ("sqrt.d %0,%1" : "=f" (z) : "f" (x));
    return z;
#else
    return __ieee754_sqrt(x);
#endif
}

/*
 * vim: ts=4 : sts=4 : sw=4 : et
 */