> > > + scancode = scancode >> 1; /* lowest bit is release bit */
> > > + down = scancode & 1;
> > Shouldn't that be the other way 'round?
> I don't know. Anyone?
Me neither, this just looked weird to me, since this code is equivalent to
down = (scancode & 2) >> 1;
scancode = scancode >> 1;
and this is not what I'd expect from the comment. :-)
Simon
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