Re: [Mesa3d-dev] GLchan and texture images

[Brian P. <br...@va...>]

Gareth Hughes wrote:
> 
> Brian Paul wrote:
> >
> > With 3.5 drivers can implement _any_ format or type of texture images they
> > want.  They just have to provide the appropriate FetchTexel() functions
> > so the software fallbacks can operate.  FetchTexel() should return GLchan
> > values.
> >
> > But yes, the s/w Mesa teximage routines should be written with the GLchan
> > type so that 16-bit and float color components will be possible.
> 
> texImage->TexFormat can't be NULL, so there has to be texure formats for
> the s/w texture images.  I'm trying to make this as clean as possible,
> as the old "Mesa formats" were based on GLubyte per channel only.
> 
> > Originally, the "Mesa formats" and texutil.[ch] were just helper routines
> > for drivers; core Mesa knew nothing about them.  It sounds like you're
> > bringing that into core Mesa.  I'm not sure of all the ramifications of
> > that.
> 
> Only the texture format stuff.  The texutil code is used by drivers to
> convert into hardware-friendly texture formats, but the texstore
> utilities need to have corresponding gl_texture_format structures for
> tightly-packed GLchan images.
> 
> Now, the gl_texture_image structure has a pointer to a gl_texture_format
> structure that defines the internal format of the image.  The
> gl_texture_format has the component sizes, bytes per texel, and custom
> FetchTexel routines that are plugged into the main gl_texture_image
> structure.

OK, this sounds good.  I see why you were asking.


> > It was compiling about a month ago.  I started testing OSMesa with 16-bit
> > color channels but didn't do any verification.   It's definitely not
> > ready for prime time yet.
> 
> Okay, I was just wondering if it was possible to test it out.  Some of
> the colormac.h macros look a little wrong, particularly this one at 16
> bits:
> 
> #define UBYTE_TO_CHAN(b)  ((GLchan) (((b) << 8) | (b)))

I think it's correct.  To convert a GLubyte color in [0,255] to a
16-bit GLchan value in [0,65535] you'd do:

  chan = b * 65535 / 255;

which is equivalent to:

  chan = (b << 8) | b;

Here's the proof:

#include <assert.h>
int main(int argc, char *argv[])
{
   int b;
   for (b = 0; b < 256; b++) {
      int c1 = b * 65535 / 255;
      int c2 = (b << 8) | b;
      assert(c1 == c2);
   }
}

:)

-Brian