Re: [PATCH v2 5/8] iommu: Switch __iommu_domain_alloc() to device ops

From: Robin Murphy
Date: Fri Jan 27 2023 - 07:16:09 EST


On 2023-01-26 23:22, Jacob Pan wrote:
Hi Robin,

On Thu, 26 Jan 2023 18:26:20 +0000, Robin Murphy <robin.murphy@xxxxxxx>
wrote:

+static int __iommu_domain_alloc_dev(struct device *dev, void *data)
+{
+ struct device **alloc_dev = data;
+
+ if (!dev_iommu_ops_valid(dev))
+ return 0;
+
+ WARN_ONCE(*alloc_dev && dev_iommu_ops(dev) !=
dev_iommu_ops(*alloc_dev),
+ "Multiple IOMMU drivers present, which the public IOMMU
API can't fully support yet. You may still need to disable one or more to
get the expected result here, sorry!\n"); +
+ *alloc_dev = dev;
+ return 0;
+}
+
struct iommu_domain *iommu_domain_alloc(struct bus_type *bus)
{
- return __iommu_domain_alloc(bus, IOMMU_DOMAIN_UNMANAGED);
+ struct device *dev = NULL;
+
+ /* We always check the whole bus, so the return value isn't
useful */
+ bus_for_each_dev(bus, NULL, &dev, __iommu_domain_alloc_dev);
+ if (!dev)
+ return NULL;
Since __iommu_domain_alloc_dev() will always return 0, bus_for_each_dev()
will never breakout until the whole dev list is iterated over. If so, would
dev only record the last one? i.e. prior results get overwritten. Maybe a
misunderstood the logic.

Yes, as the comment points out, the intent is to walk the whole bus to check it for consistency. Beyond that, we just need *a* device with IOMMU ops; it doesn't matter at all which one it is. It happens to be the last one off the list because that's what fell out of writing the fewest lines of code.

(You could argue that there's no need to repeat the full walk if the WARN_ONCE has already fired, but I'd rather keep the behaviour simple and consistent - this is only meant to be a short-term solution, and it's not a performance-critical path)

Thanks,
Robin.


+ return __iommu_domain_alloc(dev, IOMMU_DOMAIN_UNMANAGED);
}
EXPORT_SYMBOL_GPL(iommu_domain_alloc);

Thanks,

Jacob