Re: [PATCH 04/22] KVM: selftests: Compute number of extra pages needed in mmu_stress_test

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On Thu, Sep 5, 2024 at 9:42 PM Sean Christopherson <seanjc@xxxxxxxxxx> wrote:
>
> On Thu, Sep 05, 2024, James Houghton wrote:
> > On Fri, Aug 9, 2024 at 12:43 PM Sean Christopherson <seanjc@xxxxxxxxxx> wrote:
> > >
> > > Create mmu_stress_tests's VM with the correct number of extra pages needed
> > > to map all of memory in the guest.  The bug hasn't been noticed before as
> > > the test currently runs only on x86, which maps guest memory with 1GiB
> > > pages, i.e. doesn't need much memory in the guest for page tables.
> > >
> > > Signed-off-by: Sean Christopherson <seanjc@xxxxxxxxxx>
> > > ---
> > >  tools/testing/selftests/kvm/mmu_stress_test.c | 8 +++++++-
> > >  1 file changed, 7 insertions(+), 1 deletion(-)
> > >
> > > diff --git a/tools/testing/selftests/kvm/mmu_stress_test.c b/tools/testing/selftests/kvm/mmu_stress_test.c
> > > index 847da23ec1b1..5467b12f5903 100644
> > > --- a/tools/testing/selftests/kvm/mmu_stress_test.c
> > > +++ b/tools/testing/selftests/kvm/mmu_stress_test.c
> > > @@ -209,7 +209,13 @@ int main(int argc, char *argv[])
> > >         vcpus = malloc(nr_vcpus * sizeof(*vcpus));
> > >         TEST_ASSERT(vcpus, "Failed to allocate vCPU array");
> > >
> > > -       vm = vm_create_with_vcpus(nr_vcpus, guest_code, vcpus);
> > > +       vm = __vm_create_with_vcpus(VM_SHAPE_DEFAULT, nr_vcpus,
> > > +#ifdef __x86_64__
> > > +                                   max_mem / SZ_1G,
> > > +#else
> > > +                                   max_mem / vm_guest_mode_params[VM_MODE_DEFAULT].page_size,
> > > +#endif
> > > +                                   guest_code, vcpus);
> >
> > Hmm... I'm trying to square this change with the logic in
> > vm_nr_pages_required().
>
> vm_nr_pages_required() mostly operates on the number of pages that are needed to
> setup the VM, e.g. for vCPU stacks.  The one calculation that guesstimates the
> number of bytes needed, ucall_nr_pages_required(), does the same thing this code
> does: divide the number of total bytes by bytes-per-page.

Oh, yes, you're right. It's only accounting for the page tables for
the 512 pages for memslot 0. Sorry for the noise. Feel free to add:

Reviewed-by: James Houghton <jthoughton@xxxxxxxxxx>





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