On 08/01/2013 07:49 AM, Vijay Kilari wrote: > On Wed, Jul 31, 2013 at 10:44 PM, Stephen Warren <swarren at wwwdotorg.org> wrote: >> On 07/31/2013 05:37 AM, Vijay Kilari wrote: >>> On Tue, Jul 30, 2013 at 10:29 PM, Stephen Warren <swarren at wwwdotorg.org> wrote: >>>> On 07/30/2013 04:37 AM, Vijay Kilari wrote: >>>>> On Fri, Jul 26, 2013 at 10:38 PM, Stephen Warren <swarren at wwwdotorg.org> wrote: >>>> ... >>>>>> Does a kernel that's used as the crash kernel guarantee: >>>>>> >>>>>> * Never to re-use the memory that was used by the previous kernel, so >>>>>> that the spin loop code/data won't be corrupted, ever, no matter how >>>>>> long the crash recovery kernel runs. >>>>>> >>>>>> * Not use SMP, so there's never a need to re-activate the non-boot CPUs, >>>>>> which might not work if they aren't truly disabled but rather just >>>>>> running a pin loop? >>>>> >>>>> From cat /proc/iomem, normal kernel is executed from (0x80xxxxxx) with crash >>>>> kernel reserved 64M at 0xa0000000 >>>>> >>>>> 80000000-bfffffff : System RAM >>>>> 80008000-805aeddf : Kernel code >>>>> 805e2000-8063e427 : Kernel data >>>>> a0000000-a3ffffff : Crash kernel >>>>> >>>>> crash kernel is loaded to reserved memory location and is executed from there. >>>>> I could confirm this from /proc/iomem when crash kernel is running >>>>> >>>>> a0000000-a3efffff : System RAM >>>>> a0008000-a05aeddf : Kernel code >>>>> a05e2000-a063e427 : Kernel data >>>> >>>> OK, but in the crash dump kernel, is 80008000..8063e427 reserved as >>>> well, which would guarantee that the spin loop being executed by the >>>> non-crash CPUs won't be corrupted? >>> >>> The crash dump kernel runs from reserved memory area (0xa0000000 - 0xa3effffff). >>> So it should not corrupt the memory area of original kernel that was running >>> at 0x80000000,where other CPU's are in spin loop. >> >> What about dynamic allocations? > > IMHO, it is the kdump functionality to ensure that it won't corrupt > original kernel's dynamic allocations OK, if there are explicit measure to assure this already, then there's no issue.
