Messages in this thread | | | Date | Fri, 19 Oct 2007 15:53:12 -0400 | From | Phillip Susi <> | Subject | Re: [PATCH 0/2 -mm] kexec based hibernation -v5 |
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Huang, Ying wrote: > The hibernation procedure with the patch set is as follow: > > 1. Boot a kernel A > > 2. Work under kernel A > > 3. Kexec another kernel B (crash dump enabled) in kernel A. > > 4. Save the memory image of kernel A through crash dump (such as "cp > /proc/vmcore ~"). Save the "jump back entry".
Doesn't this also save the memory of kernel B?
> 5. Shutdown or reboot > > > The restore process with the patch set is as follow: > > 1. Boot a kernel C (crash dump enabled), the memory area used by > kernel C must be a subset of memory area used by kernel B.
Why is a third kernel needed? Why can't kernel B be used for this as well? In fact, if kernel A has been compiled to be relocatable and crash dump enabled, why wouldn't it suffice for all 3 instances?
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