On Fri, Apr 29, 2022 at 9:59 AM Paolo Bonzini <pbonzini@xxxxxxxxxx> wrote:
>
> On 4/29/22 17:51, Peter Gonda wrote:
> >> No, you don't need any of this. You can rely on there being only one
> >> depmap, otherwise you wouldn't need the mock releases and acquires at
> >> all. Also the unlocking order does not matter for deadlocks, only the
> >> locking order does. You're overdoing it. :)
> >
> > Hmm I'm slightly confused here then. If I take your original suggestion of:
> >
> > bool acquired = false;
> > kvm_for_each_vcpu(...) {
> > if (acquired)
> > mutex_release(&vcpu->mutex.dep_map,
> > _THIS_IP_); <-- Warning here
> > if (mutex_lock_killable_nested(&vcpu->mutex, role)
> > goto out_unlock;
> > acquired = true;
> >
> > """
> > [ 2810.088982] =====================================
> > [ 2810.093687] WARNING: bad unlock balance detected!
> > [ 2810.098388] 5.17.0-dbg-DEV #5 Tainted: G O
> > [ 2810.103788] -------------------------------------
>
> Ah even if the contents of the dep_map are the same for all locks, it
> also uses the *pointer* to the dep_map to track (class, subclass) ->
> pointer and checks for a match.
>
> So yeah, prev_cpu is needed. The unlock ordering OTOH is irrelevant so
> you don't need to visit the xarray backwards.
Sounds good. Instead of doing this prev_vcpu solution we could just
keep the 1st vcpu for source and target. I think this could work since
all the vcpu->mutex.dep_maps do not point to the same string.
Lock:
bool acquired = false;
kvm_for_each_vcpu(...) {
if (mutex_lock_killable_nested(&vcpu->mutex, role)
goto out_unlock;
acquired = true;
if (acquired)
mutex_release(&vcpu->mutex, role)
}
Unlock
bool acquired = true;
kvm_for_each_vcpu(...) {
if (!acquired)
mutex_acquire(&vcpu->mutex, role)
mutex_unlock(&vcpu->mutex);
acquired = false;
}
So when locking we release all but the first dep_maps. Then when
unlocking we acquire all but the first dep_maps. Thoughts?
>
> Paolo
>
