Re: Linux non-root node question

2019-06-24 Thread Wayne
... 0x3ff/8] = 0, /* serial1 */ }, Also, is there any way to prevent the non-root from trying to probe against ttyS0 at startup and causing a memory access violation? I'm aware of the 8250.nr_uarts=1, but it seems to default to enumerating ttyS0. Thanks again for all of your help, Wayne On Wed, J

Re: Linux non-root node question

2019-06-19 Thread Wayne
the filesystem being entirely in the ramdisk. Any thoughts on how to accomplish this? It seems like buildroot can generate images based on busybox, with the ash shell. I really would like bash support. Thanks for all of your help getting to this point, Wayne On Tue, Jun 18, 2019 at 1:09 PM

Re: Linux non-root node question

2019-06-18 Thread Wayne
, and also jailhouse fails to create the guest if I attempt to allocate more memory. 3. I can get to the kernel login prompt (and login successfully) if I use your rootfs.cpio binary. This option so far is the most promising. Thanks for your assistance, Wayne On Tue, Jun 18, 2019 at 12:01 PM

Re: Linux non-root node question

2019-06-18 Thread Wayne
ramfs files. Also, the issue where my serial output scrolled continuously was due to the capture method. I switched to minicom and no longer see that occur. Any further thoughts on why jailhouse fails to load the guest when I set the memory region greater than 200MB? Thanks, Wayne On Tue, J

Re: Linux non-root node question

2019-06-18 Thread Wayne
ort "optional"): VT-x (VMX) : VMX inside SMX missing (optional) VT-d (IOMMU #0-3) : 39-bit AGAW missing (optional) (2M pages and 1G pages show as ok) Wayne On Tue, Jun 18, 2019 at 2:32 AM Jan Kiszka wrote: > On 17.06.19 21:06, Wayne wrote: > > Hi Jan, > > > > I a

Re: Linux non-root node question

2019-06-17 Thread Wayne
uld match the 4.19 guest. Note that my root kernel is vanilla 4.16 and my non-root linux guest is 4.19 jailhouse enabling from siemens. I attached my latest System config and non-linux cell config. Thanks for your repeated help, Wayne On Thu, Jun 13, 2019 at 2:55 PM Jan Kiszka wrote: > O

Re: Linux non-root node question

2019-06-13 Thread Wayne
config. Should I expect to be able to log in to the non-root if I use the same initramfs as the root linux? Wayne On Thu, Jun 13, 2019 at 2:06 PM Jan Kiszka wrote: > On 13.06.19 20:01, Wayne wrote: > > Jan/Ralf: > > > > I tried your suggestion with using the root node i

Re: Linux non-root node question

2019-06-12 Thread Wayne
uot; command take an initramfs image (.img) through the "-i option" or does it need to be an initrd? If initrd only, can you point me to an example x86 rootfs.cpio image? I'm having trouble installing Buildroot on my machine. Thanks again, Wayne On Wed, Jun 12, 2019 at 1:17 P

Re: Linux non-root node question

2019-06-12 Thread Wayne
ooting the non-root besides the serial concern, I haven't provided an initramfs to my non-root linux yet. I would like to probably boot it with a simple ramdisk if possible to get a point where I can log in. Thanks, Wayne On Wed, Jun 12, 2019 at 6:43 AM Ralf Ramsauer < ralf.ramsa...@oth-reg

Re: Linux non-root node question

2019-06-11 Thread Wayne
port: 3f9: size 1" Is it possible to map the same pio block in both the root and non-root with a different mask? Or does the non-root just override it? It seems like the latter is true. Thanks again Jan and Ralf for your help getting to this point. On Tue, Jun 11, 2019 at 2:13 PM

Re: Linux non-root node question

2019-06-11 Thread Wayne
r bare minimum x86 kernel config on your machine for the guest to get around these? I know Jan mentioned the jailhouse-images project might have something, but I couldn't find it digging around quick. Thanks to you both for your responses and patience. I got the root linux up and going on my own

Linux non-root node question

2019-06-07 Thread Wayne
ce/jailhouse-dev/M7UO89XFIk0/Qi40DDuMBAAJ ". Any information you can provide to me will be helpful. I'm not sure what might be going wrong here. Thanks, Wayne -- You received this message because you are subscribed to the Google Groups "Jailhouse" group. To unsubscribe