AWARE SYSTEMS
TIFF and LibTiff Mail List Archive

Thread

2004.05.20 11:05 "[Tiff] Extra samples handling", by Mayank
2004.05.20 23:42 "Re: [Tiff] Extra samples handling", by Chris Cox
2004.05.21 09:05 "Re: [Tiff] Extra samples handling", by Mayank
2004.05.21 19:22 "Re: [Tiff] Extra samples handling", by Chris Cox
2004.05.21 12:00 "[Tiff] Un-premultiplying the color data", by Mayank
2004.05.21 19:30 "[Tiff] Re: Un-premultiplying the color data", by Chris Cox

2004.05.21 19:30 "[Tiff] Re: Un-premultiplying the color data", by Chris Cox

Here is what I am trying in my code to un-premultiply the color data in the image attached with this message:

source: is the buffer returned by the libtiff functions.
dest: is the destination buffer
the image is: contig, strip, CMYK, 8-bits per channel, with one associated alpha channel

since i have to gather the color values in my dest buffer in the order KCMY, that is why i take dest[0] = source[3] and so on.

alpha = source[4];
if(alpha != 0)
{
dest[0] = ( (source[3]*255) + alpha/2 )/alpha;
dest[1] = ( (source[0]*255) + alpha/2 )/alpha;
dest[2] = ( (source[1]*255) + alpha/2 )/alpha;
dest[3] = ( (source[2]*255) + alpha/2 )/alpha;
}

It would run faster is you used a lookup table for the divide and multiply by 255.

Divides are slow on all modern CPUs, and even a 64k lookup table is faster -- unless you've got someone who can write really good SSE or AltiVec code.

Also, using a lookup table would avoid divides by zero, and you can special case top == 0 (which should return zero) and bottom <= top (which should return 255) without any branches.

Also, CMYK is white matted - what you're doing is black matting and only suitable for RGB data.

To white matte:

        out_matted = 255 - (alpha * in + 127) / 255;

To un-white matte:

        out = 255 - ( (255* (255 - in_matted) + (alpha/2)) / alpha );

There are two questions here:

  1. How do you know come to know about the color information, if alpha had been zero?

You don't. If the area is fully transparent, there is no color information.

  1. Is the procedure above correct?

Nope.

Chris