running into OOM killer with kexec loading large ramdisk

running into OOM killer with kexec loading large ramdisk

[Tobias Powalowski]

Hi,
again me,
I try to load a 900MB ramdisk on a ramfs rootfs  with kexec and a 10MB kernel.
With 2800MB RAM assigned to qemu.
Memory free by /proc/memstat: 2.2GB
It keeps on OOM killed while executing:
kexec -l kernel --initrd=initrd.img
I can safely unpack the initrd in the ramfs without getting OOM killed.
What is kexec doing wrong here?
greetings
tpowa


-- 
Tobias Powalowski
Arch Linux Developer & Package Maintainer (tpowa)
https://www.archlinux.org
tpowa at archlinux.org

St. Martin-Apotheke
Herzog-Georg-Str. 25
89415 Lauingen
https://www.st-martin-apo.de
info at st-martin-apo.de

running into OOM killer with kexec loading large ramdisk

[Philipp Rudo]

Hi Tobias,

On Wed, 23 Mar 2022 12:14:59 +0100
Tobias Powalowski <tobias.powalowski@googlemail.com> wrote:

> Hi,
> again me,
> I try to load a 900MB ramdisk on a ramfs rootfs  with kexec and a 10MB kernel.
> With 2800MB RAM assigned to qemu.
> Memory free by /proc/memstat: 2.2GB
> It keeps on OOM killed while executing:
> kexec -l kernel --initrd=initrd.img
> I can safely unpack the initrd in the ramfs without getting OOM killed.
> What is kexec doing wrong here?

I don't think that kexec is doing anything wrong here.

The kexec_load syscall is designed in a way that user space prepares
everything in a huge buffer and passes a pointer to it to the system
call. The systemcall then needs to copy everything from the user buffer
to a kernel buffer. So there are three copies of the initrd in memory
(including the one in ramfs). Together they take up 3 * 900MB = 2.7GB.
So basically all the memory of your guest.

You can try using the kexec_file_load systemcall. That at least
eliminates the copy in the user buffer so there are only two copies
left. 400MB of free memory is still quite scarce, though.

Thanks
Philipp

running into OOM killer with kexec loading large ramdisk

[Tobias Powalowski]

Hi, thanks for the explanation,
I got it now working, with running the kexec call in the background
and deleting the initramfs in the foreground after sleep 1.
I still think kexec should handle this with less memory needed.
greetings
tpowa

Am Fr., 25. M?rz 2022 um 13:26 Uhr schrieb Philipp Rudo <prudo@redhat.com>:
>
> Hi Tobias,
>
> On Wed, 23 Mar 2022 12:14:59 +0100
> Tobias Powalowski <tobias.powalowski@googlemail.com> wrote:
>
> > Hi,
> > again me,
> > I try to load a 900MB ramdisk on a ramfs rootfs  with kexec and a 10MB kernel.
> > With 2800MB RAM assigned to qemu.
> > Memory free by /proc/memstat: 2.2GB
> > It keeps on OOM killed while executing:
> > kexec -l kernel --initrd=initrd.img
> > I can safely unpack the initrd in the ramfs without getting OOM killed.
> > What is kexec doing wrong here?
>
> I don't think that kexec is doing anything wrong here.
>
> The kexec_load syscall is designed in a way that user space prepares
> everything in a huge buffer and passes a pointer to it to the system
> call. The systemcall then needs to copy everything from the user buffer
> to a kernel buffer. So there are three copies of the initrd in memory
> (including the one in ramfs). Together they take up 3 * 900MB = 2.7GB.
> So basically all the memory of your guest.
>
> You can try using the kexec_file_load systemcall. That at least
> eliminates the copy in the user buffer so there are only two copies
> left. 400MB of free memory is still quite scarce, though.
>
> Thanks
> Philipp
>


-- 
Tobias Powalowski
Arch Linux Developer & Package Maintainer (tpowa)
https://www.archlinux.org
tpowa at archlinux.org

St. Martin-Apotheke
Herzog-Georg-Str. 25
89415 Lauingen
https://www.st-martin-apo.de
info at st-martin-apo.de