On Saturday, 14 July 2007 07:48, Huang, Ying wrote:
> On Fri, 2007-07-13 at 10:43 -0600, Eric W. Biederman wrote:
> > > Why a assembly stub is necessary? Is it not sufficient that just
> > > continue to complete a normal boot (hot add the reset of memory) or load
> > > the hibernated kernel (hibernated image) and jump to it?
> >
> > I was thinking the assembly stub would be the small piece that jumps
> > to loaded hibernated kernel. Quite possibly we could just get away
> > with providing no memory and just an entry point to kexec but it
> > makes sense to me to plan on running a couple of instructions.
>
> Oh, I got it. In my patch, there is such assembly stub in
> arch/i386/kernel/kexec_jump.S. I think it is needed to restore basic CPU
> state and accommodate some position independent restore code (such as
> memory restore code).
>
> > Actually the way the kexec infrastructure it might be reasonable to
> > just use sys_kexec_load to load the entire hibernated image. Except
> > for the fact that sys_kexec_load requires the source pages to be
> > in the processes memory image the code shouldn't have the 50% of
> > memory limitation already.
> >
> > If we can get that going we don't even need to restrict the first
> > kernels memory. So it might just require teaching sys_kexec_load
> > how to steal process pages. Anyway something to think about.
>
> As for memory backupping and restoring during hibernating and resuming,
> I think a possible picture can be as follow:
>
> Memory:
>
> Total memory: 512M
> Memory used by hibernating/resuming kernel: 0~16M
>
>
> Hibernating process:
>
> 1. Normal kernel running
> 2. Hibernating is triggered, sys_kexec_load is used to load
> hibernating kernel and initramfs into memory. Then
> sys_reboot(LINUX_REBOOT_CMD_KSPAWN) is invoked.
> 3. In sys_reboot, kexec_jump is called to save device/CPU state,
> then relocate_kernel is called. kexec_jump and relocate_kernel
> reside in individual page in 16M~512M.
OK
What's going to happen to devices at this point?
> 4. In relocate_kernel, 0~16M is backupped firstly, then the
> hibernating kernel and initramfs is copied to 0~16M, after that,
> the hibernating kernel is booted.
> 5. In hibernating kernel, the memory of normal kernel (it is in
> 16M~512M) is saved into a hibernation image through /dev/mem
> and ELF header.
I don't think it can be _that_ simple:
(a) what about processes' memory
(b) what about areas that shouldn't be saved?
> Resume process:
>
> 1. Resuming kernel is booted as a normal kernel, but the memory is
> restricted to 0~16M.
> 2. Checking whether there is a effective hibernation image. If
> there isn't, the memory of 16M~512M is hot added, and the normal
> boot up process continues; If there is, a resuming process is
> triggered.
> 3. sys_kexec_load is used to restore the memory state of hibernated
> kernel. The sys_kexec_load works in crashdump way, that is, the
> hibernation image is copied to destination location in 16M~512M
> in sys_kexec_load instead of relocate_kernel. There is no half
> of memory size restriction.
> 4. sys_reboot is called to trigger jumping back, which will jump back
> to kexec_jump of hibernated kernel.
> 5. In kexec_jump of hibernated kernel, the memory of 0~16M is copied
> back from the backup area in 16M~512M. The memory state of
> hibernated kernel is restored totally. The CPU and device state
> can be restored after that.
Well, I don't know why this needs to be that complicated. We already have
code in the mainline that's able to load a large hibernation image into memory
and jump to the kernel being restored. And it has _no_ 50% of RAM limitation,
this is the _saving_ part of the current code that this limitation comes from.
Greetings,
Rafael
--
"Premature optimization is the root of all evil." - Donald Knuth
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