Re: Off by eight.

[Jordan Archer]

How does the fallback version get executed?  Stepping through the reset 
sequence it looks from my image that that's the default.

Jordan

At 02:52 PM 11/18/2003, ron minnich wrote:
> On Tue, 18 Nov 2003, Jordan Archer wrote:
> > I'm pretty sure that it's not 32 bit reset code.  The problem appears if I
> > try to build an image without a fallback.
>
> oh. When I try to build an image without a fallback (this part is
> counterintuitive) I always just build a fallback. Possibly that's the
> wrong thing to do :0)
>
> ron