On Wed, Jul 22, 2009 at 04:39:55PM +0000, Omari Stephens wrote:
> Frank van Maarseveen wrote:
> > On Wed, Jul 22, 2009 at 08:21:29AM +0100, Martin Ling wrote:
> >> On Wed, Jul 22, 2009 at 04:26:45AM +0000, Omari Stephens wrote:
> >>>> data->unnormalized_angle =
> >>>> fmod((data->unnormalized_angle + 180.0) / 360.0) - 180.0;
> >>> Do you mean fmod(data->unnormalized_angle + 180.0, 360.0) - 180.0; ?
> >>> fmod takes
> >>> two arguments.
> >> Yes. ;-)
> >
> > I've played with the idea too but it won't work. modulo 360 will yield
> > -360..360 which is a range of 720 degrees, not 360. Modulo 180 yields
> > the correct range but then the result is incorrect.
>
> Yeah, the problem here is that fmod (and all "mod" operators/functions in
> anything C-like) actually implements the remainder operation and not the
> modulo
> operation. Basically, that means that the output range of fmod(x, N) is [-N,
> N)
> rather than [0, N) with the modulo operator.
>
> This is one reason I hate C, and every other language that figured it'd be
> good
> to be compatible with C :o) (Note that I have no idea if Fortran behaves
> this
> way or not 8)
Well, you just hinted me at the (C99) solution :-P
data->unnormalized_angle = remainder(data->unnormalized_angle, 360);
x fmod(x,360) remainder(x,360)
170.000000: 170.000000 170.000000
190.000000: 190.000000 -170.000000
-170.000000: -170.000000 -170.000000
-190.000000: -190.000000 170.000000
350.000000: 350.000000 -10.000000
-350.000000: -350.000000 10.000000
370.000000: 10.000000 10.000000
-370.000000: -10.000000 -10.000000
test program:
#include <stdio.h>
#include <math.h>
void main(void)
{
double f;
while (scanf("%lf", &f) == 1)
printf("%f: %f %f\n", f, fmod(f, 360), remainder(f,
360));
}
fmod(x, y):
The return value is x - n * y, where n is the quotient of x / y,
rounded towards zero to an integer.
remainder(x, y):
The return value is x-n*y, where n is the value x / y, rounded
to the nearest integer.
notice the difference: towards zero vs. nearest integer.
It's not that important but I couldn't resist this puzzle :-)
--
Frank
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