Am 06.02.03, 19:47 +0100 schrieb Mart� Maria:
...
> >I use:
> > f32Lab[0] = src[0] * 100; // CIE L
> > f32Lab[1] = (src[1] - 0.5 ) * 256 ; // CIE a
> > f32Lab[2] = (src[2] - 0.5 ) * 256 ; // CIE b
> > cmsDoTransform(hTrans_f32lab_u8rgb, f32Lab, u8RGB, 1); .
The encoding of src is double with an range of 0.0 - 1.0 for each of the
three samples. My most interesst is using floats with at least 16bit.
>
> Humm... what is the encoding of src[]? Is this TIFF Lab? If so,
> you don't need floating point. This will slow down whole transform.
> Just open lcmstiff8.icm, use TYPE_Lab_8 and then apply transform
> to raw data.
>
> // ----
> hLab = cmsOpenProfileFromFile("lcmstiff8.icm", "r");
> sRGB = cmsCreate_sRGBProfile();
> hTrans_f32lab_rgb = cmsCreateTransform(hLab, TYPE_Lab_8,
I have doules here-^
> hsRGB, TYPE_RGB_8,
> INTENT_PERCEPTUAL, 0);
>
> cmsDoTransform(hTrans_f32lab_u8rgb, src, u8RGB, 1); .
> // ------
>
> This will be way faster since no floating point conversions are involved.
Awaiting ;-)
>
> If what you want is really convert floating point values, then you could
> use TYPE_Lab_DBL and pass the Lab as an array of doubles. In both
> cases you need not the "f32Lab[1] = (src[1] - 0.5 ) * 256 " part.
>
> // ------
> hLab = cmsCreateLabProfile(NULL);
> sRGB = cmsCreate_sRGBProfile();
> hTrans_f32lab_rgb = cmsCreateTransform(hLab, TYPE_Lab_8,
> hsRGB, TYPE_RGB_8,
> INTENT_PERCEPTUAL, 0);
I dont understand how it makes sense to switch from doubles to 8bit
integers for hLab? You really mean downsampling?
> double Lab[3];
>
> Lab[0] = 50; // The Lab value Lab=(50, -10.2, 23.6)
> Lab[1] = -10.2;
> Lab[2] = 23.6;
>
> cmsDoTransform(hTrans_f32lab_u8rgb, Lab, u8RGB, 1); .
> // ------
I think I did it with the my above code. Still, the result is not what I
expected. (I used f32 as simple float and with casting to double -> same
result.)
Maybe as You don't use floats anyway, I rewrite it and use 16bit integers.
That worked right.
> Regards,
> Mart�.
thanks
Kai-Uwe
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