Re: [PATCH 2/3] iommu/vt-d: Clear Present bit before tearing down PASID entry

[Baolu Lu <baolu.lu@linux.intel.com>]

On 1/16/26 05:35, Dmytro Maluka wrote:
> On Thu, Jan 15, 2026 at 10:45:12AM +0800, Baolu Lu wrote:
>> On 1/14/26 19:12, Dmytro Maluka wrote:
>>> On Wed, Jan 14, 2026 at 01:38:13PM +0800, Baolu Lu wrote:
>>>> On 1/14/26 03:34, Dmytro Maluka wrote:
>>>>> On Tue, Jan 13, 2026 at 11:00:47AM +0800, Lu Baolu wrote:
>>>>>> +	intel_pasid_clear_entry(iommu, dev, pasid, fault_ignore);
>>>>> Is it safe to do this with iommu->lock already unlocked?
>>>>
>>>> Yes, it is. The PASID entry lifecycle is serialized by the iommu_group-
>>>>> mutex in the iommu core, which ensures that no other thread can attempt
>>>> to allocate or setup this same PASID until intel_pasid_tear_down_entry()
>>>> has returned.
>>>>
>>>> The iommu->lock is held during the initial transition (P->0) to ensure
>>>> atomicity against other hardware-table walkers, but once the P bit is
>>>> cleared and the caches are flushed, the final zeroing of the 'dead'
>>>> entry does not strictly require the spinlock because the PASID remains
>>>> reserved in software until the function completes.
>>>
>>> Ok. Just to understand: "other hardware-table walkers" means some
>>> software walkers, not hardware ones? Which software walkers are those?
>>> (I can't imagine how holding a spinlock could prevent the hardware from
>>> walking those tables. :))
>>
>> You are right. A spinlock doesn't stop the hardware. The spinlock
>> serializes software threads to ensure the hardware walker always sees a
>> consistent entry.
>>
>> When a PASID entry is active (P=1), other kernel paths might modify
>> the control bits in-place. For example:
>>
>> void intel_pasid_setup_page_snoop_control(struct intel_iommu *iommu,
>>                                            struct device *dev, u32 pasid)
>> {
>>          struct pasid_entry *pte;
>>          u16 did;
>>
>>          spin_lock(&iommu->lock);
>>          pte = intel_pasid_get_entry(dev, pasid);
>>          if (WARN_ON(!pte || !pasid_pte_is_present(pte))) {
>>                  spin_unlock(&iommu->lock);
>>                  return;
>>          }
>>
>>          pasid_set_pgsnp(pte);
>>          did = pasid_get_domain_id(pte);
>>          spin_unlock(&iommu->lock);
>>
>>          intel_pasid_flush_present(iommu, dev, pasid, did, pte);
>> }
>>
>> In this case, the iommu->lock ensures that if two threads try to modify
>> the same active entry, they don't interfere with each other and leave
>> the entry in a 'torn' state for the IOMMU hardware to read.
>>
>> In intel_pasid_tear_down_entry(), once the PASID entry is deactivated
>> (setting P=0 and flushing caches), the entry is owned exclusively  by
>> the teardown thread until it is re-configured. That's the reason why the
>> final zeroing doesn't need the spinlock.
> 
> I see. Am I correct that those other code paths (modifying an entry
> in-place) are not supposed to do that concurrently with
> intel_pasid_tear_down_entry(), i.e. they should only do that while it is
> guaranteed that the entry remains present? Otherwise there is a bug
> (hence, for example, the WARN_ON in
> intel_pasid_setup_page_snoop_control())?

The iommu driver assumes that high-level software should ensure this.

> So, holding iommu->lock during
> entry teardown is not strictly necessary (i.e. we could unlock it even
> earlier than setting P=0), i.e. holding the lock until the entry is
> deactivated is basically just a safety measure for possible buggy code?

There are other paths that may be concurrent, such as the debugfs path
(dumping the pasid table through debugfs). Therefore, keeping iommu-
 >lock in the driver is neither redundant nor buggy.

Thanks,
baolu