general protection error:0
mikey at neuling.org
Thu Apr 29 18:04:03 EDT 2010
> I'm running Ubuntu Hardy with a 2.6.28 kernel on VirtualBox, and want to
> simply restart the current kernel (because I'm a kexec noob and that
> seems a sensible first step.) I've put "break=init" on the command line
> so it gives me a shell just before pivoting root from the initrd to the
> hard-disk. I do the kexec -l and kexec -e, the kernel (re)starts, does
> all of the usual startup things, and drops me back in a shell when it's
> about to start init (break=init, remember?) At this point, when I try
> to run almost any command, for example "chroot /root bash", I get
> "Segmentation fault" with general protection logged in dmesg, for
> example: "bash general protection ip:b7f0c425 sb:bfb282f4
> error:0" No doubt I'm making a basic mistake, but Google failed to show
> me what, hence I'm asking here for suggestions.
> Has anyone any suggestion?
2.6.28 is pretty old these days. It's unlikely someone here is going to
help you debug a kernel that ancient.
Can you retry with current mainline and kexec-tools user space? Also
post your full boot log and kexec -l and -e outputs when you hit this
More information about the kexec