2007/11/10, Sven Neumann <[EMAIL PROTECTED]>:

> On Sat, 2007-11-10 at 19:25 +0100, Giacomo Mazzocato wrote:
> > I've tried to reimplement it in another language
> I very much hope for you that you are respecting the license that this
> code has been published under. Your code is GPL, right?

Now it's for private use and I if I release it i will be GPL.
But can you put a license on an algorithm?

> and in the function
> > cdisplay_colorblind_convert with certain colors (for instance #000000) I
> get a
> > warning of division by zero in the section that corresponds to this in
> the
> > original code:
> >  switch (colorblind->deficiency)
> >           {
> >               tmp = blue / red;
> >               ^^^^^^^^^^
> Division by zero is handled correctly here. The result will be NaN.

I'll do further checking on how Nan is handled as soon as I can lay a hand
on a C compiler. In particular I wonder if Nan can be less than a normal
numberic value because the lines that follow in that function depend on

> The most important problem is however that when I try to convert white
> > (#ffffff) with the filter,I get #dadada for all 3 deficit simulations.
> >
> > #dadada is a shade of grey, but if I apply in gimp the filter to an area
> which
> > contains white regions, white remains white.
> If you have other display filters active, then the colors may be passed
> through them befor they are displayed. Otherwise, no, there is no
> further post-processing being done.

No other filter is active but I still see white if I convert a white region.
That's strange. Can you confirm that #ffffff is converted to #dadada ?

Perhaps you should just link with libcolorblind (see
> http://colorblind.alioth.debian.org/) instead of reimplementing the
> algorithm...

I'll check that library, thanks for the link
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