On Mon, Aug 24, 2009 at 12:47:42PM -0500, Kevin Grittner wrote:
> David Fetter <david@fetter.org> wrote:
> > On Mon, Aug 24, 2009 at 11:14:19PM +1000, Paul Matthews wrote:
>
> > These next two lines are a teensy bit baroque. Is there some
> > significant speed increase that would justify them?
> >
> >> if (x == 0.0)
> >> return 0.0;
> >> else {
> >> yx = y/x;
> >> return x*sqrt(1.0+yx*yx);
> >> }
> >> }
>
> I think the reason is overflow. From the function comment:
>
> >> * The traditional formulae of x^2+y^2 is rearranged
> >> * to bring x outside the sqrt. This allows computation of the
> hypotenuse
> >> * for much larger magnitudes than otherwise normally possible.
>
> Although I don't see why the first part isn't:
>
> if (y == 0.0)
> return x;
D'oh!
Good point :)
So the code should read as follows?
#include <math.h>
#include "c.h"
#include "utils/builtins.h"
/** Find the hypotenuse. Firstly x and y are swapped, if required, to make* x the larger number. The traditional
formulaeof x^2+y^2 is rearranged* to bring x outside the sqrt. This allows computation of the hypotenuse* for much
largermagnitudes than otherwise normally possible.** sqrt( x^2 + y^2 ) = sqrt( x^2( 1 + y^2/x^2) )* =
x* sqrt( 1 + y^2/x^2 )* = x * sqrt( 1 + y/x * y/x )*/
double hypot( double x, double y )
{ double yx;
if( isinf(x) || isinf(y) ) return get_float8_infinity();
if( isnan(x) || isnan(y) ) return get_float8_nan();
x = fabs(x); y = fabs(y);
if (x < y) { double temp = x; x = y; y = temp; } if (y == 0.0) return x; yx = y/x;
returnx*sqrt(1.0+yx*yx);
}
Cheers,
David.
--
David Fetter <david@fetter.org> http://fetter.org/
Phone: +1 415 235 3778 AIM: dfetter666 Yahoo!: dfetter
Skype: davidfetter XMPP: david.fetter@gmail.com
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