On 06.08.20 23:32, Stefano Stabellini wrote:
Hi Stefano
On Thu, 6 Aug 2020, Jan Beulich wrote:
On 06.08.2020 02:37, Stefano Stabellini wrote:
What should do_trap_stage2_abort_guest do on IO_RETRY? Simply return
early and let the scheduler do its job? Something like:
enum io_state state = try_handle_mmio(regs, hsr, gpa);
switch ( state )
{
case IO_ABORT:
goto inject_abt;
case IO_HANDLED:
advance_pc(regs, hsr);
return;
case IO_RETRY:
/* finish later */
return;
case IO_UNHANDLED:
/* IO unhandled, try another way to handle it. */
break;
default:
ASSERT_UNREACHABLE();
}
Then, xen/arch/arm/ioreq.c:handle_mmio() gets called by
handle_hvm_io_completion() after QEMU completes the emulation. Today,
handle_mmio just sets the user register with the read value.
But it would be better if it called again the original function
do_trap_stage2_abort_guest to actually retry the original operation.
This time do_trap_stage2_abort_guest calls try_handle_mmio() and gets
IO_HANDLED instead of IO_RETRY, thus, it will advance_pc (the program
counter) completing the handling of this instruction.
The user register with the read value could be set by try_handle_mmio if
try_fwd_ioserv returns IO_HANDLED instead of IO_RETRY.
Is that how the state machine is expected to work?
I think so. Just because it has taken us quite some time (years) on
the x86 side to get reasonably close to how hardware would behave
(I think we're still not fully there): The re-execution path needs
to make sure it observes exactly the same machine state as the
original path did. In particular changes to memory (by another vCPU)
must not be observed.
Thanks for the heads up. I think I understand how it is supposed to work
now. I hope Oleksandr is on the same page.
Not completely. I am still going through the discussion and navigating
the code trying to understand missing bits for me.
Thanks for trying to clarify how it supposed to work.
--
Regards,
Oleksandr Tyshchenko
