On Mon, 2003-02-10 at 18:55, Alexandr Andreev wrote:
> Michel Dänzer wrote:
> > On Fre, 2003-02-07 at 21:36, Alexandr Andreev wrote:
> >>
> >>Bool
> >>XAACheckStippleReducibility(PixmapPtr pPixmap)
> >>{
> >>...
> >>pPriv->pattern0 = bits[0] | SHIFT_L(bits[1],8) | SHIFT_L(bits[2],16) |
> >>SHIFT_L(bits[3],24);
> >>pPriv->pattern1 = bits[4] | SHIFT_L(bits[5],8) | SHIFT_L(bits[6],16) |
> >>SHIFT_L(bits[7],24);
> >>...
> >>}
> >>where SHIFT_L(value, shift) is defined as ((value) >> (shift)) for Big
> >>Endian.
> >>
>
> SHIFT_L really means shift right for big endian machines!
>
> >>Bool
> >>XAACheckTileReducibility(PixmapPtr pPixmap, Bool checkMono)
> >>{
> >>...
> >>pPriv->pattern0 = bits[0] | (bits[1]<<8) | (bits[2]<<16) | (bits[3]<<24);
> >>pPriv->pattern1 = bits[4] | (bits[5]<<8) | (bits[6]<<16) | (bits[7]<<24);
> >>...
> >>}
> >>
>
> Here is the true shift left.
>
> >>In both cases the unsigned int bits[] array contains bytes!
> >
> >
> > No, in XAACheckStippleReducibility() it's declared as CARD32.
> >
>
> Yes of course, but In the first case the __most__ significant byte of
> the bits[0] contains data and in the second case the __last__ significant
> byte of the bits[0] contains data. So, you obtain swapped pattern0 and
> pattern1 in the "tile" case.
I see where you're getting at, but does bits[0] really have the same
meaning in both cases?
--
Earthling Michel Dänzer (MrCooper)/ Debian GNU/Linux (powerpc) developer
XFree86 and DRI project member / CS student, Free Software enthusiast
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