Hello Julien, As requested I am moving the conversation to email from IRC
To summarize my setup: 1. I am running a custom kernel on QEMU ARM64(without KVM) on my linux machine 2. I have my custom implementation of Hypervisor 3. I am trying to run the same custom kernel as guest OS on top of my Hypervisor - I am able to boot my kernel to shell on QEMU - I am able to start my guest OS - From the logs I see that my guest OS finishes booting up, I can see the $sign for the shell and then it goes into idle state, but I cannot use the shell To debug further I enabled tracing in QEMU and printed the exceptions to understand what state is my guest in Before I paste some logs here, some more information about my system IRQ 30 is the physical timer interrupt of my host OS running on QEMU IRQ 27 is the virtual timer interrupt of my guest OS I have added some extra logging in QEMU to print out the VTTBR so as to understand where the exception is coming from >From 1st Attachment (GIC 1) I observe that every once in a while the phys timer interrupt occurs (IRQ 30) and its handled by the host VM, and then after about 10-20 times the virtirq 27 level changes to 1 and back to 0 again and again..this is how its looping currently after boot up Adding a 2nd attachment with extra logging of traps of exceptions as well. It just shows 2 different IRQ exceptions taken, one with VTTBR = 0 (IRQ30) and other with a VTTBR value of the guest (IRQ 27, since the irq 27 level is changed to 1 just before it.. Any idea what am I missing, and why my guest OS is not handling the pending interrupt. Upon receiving the IRQ 27, I do set the HCR_EL2.VI bit to 1 to signal the guest about a pending virual interrupt but I dont think thats working. Another thing I noticed that the qemu logging, never shows anything for the virt_interrupt. Appreciate any pointers to help debug this. Thanks, BR
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