Re: correctly rounded sqrt

[Szabolcs Nagy <nsz@port70.net>]

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* Szabolcs Nagy <nsz@port70.net> [2012-03-14 20:01:23 +0100]:
> * Szabolcs Nagy <nsz@port70.net> [2012-03-14 17:56:05 +0100]:
> > i tried to solve the sqrt issue with a wrapper

ok as discussed a simpler and better solution
without precision setting etc in c

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#include <math.h>
#include <stdint.h>

/* dekker exact mult u*u == hi + lo */
static void sq(long double *hi, long double *lo, long double u)
{
	static const long double c = 1.0 + 0x1p33;
	long double cu, u1, u2;

	cu = c*u;
	u1 = (u - cu) + cu;
	u2 = u - u1;
	*hi = u*u;
	*lo = (u1*u1 - *hi) + u1*u2*2.0 + u2*u2;
}

/* correctly rounded double sqrt using long double arithmetics for x87 */
double x_sqrt(double x)
{
	union {
		long double x;
		struct{
			uint64_t m;
			uint16_t es;
			uint16_t pad;
		} bits;
	} u;
	long double r, hi, lo;

	u.x = sqrtl(x);
	if ((u.bits.m&0x7ff) == 0x400) {
		if ((u.bits.es&0x7fff) == 0x7fff)
			return u.x;
		sq(&hi, &lo, u.x);
		r = x - hi - lo;
		if (r > 0)
			u.bits.m++;
		else
			u.bits.m--;
		return u.x;
	}
	return u.x;
}