general protection error:0

Michael Neuling mikey at
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[1058] 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


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