On Thu, Jul 27, 2017 at 09:13:42AM +0000, Shameerali Kolothum Thodi wrote:
[...]
> > > >> +int iort_iommu_its_get_resv_regions(struct device *dev, struct
> > list_head *head)
> > > >> +{
> > > >> + int i;
> > > >> + struct acpi_iort_its_group *its;
> > > >> + struct acpi_iort_node *node, *its_node = NULL;
> > > >> + int resv = 0;
> > > >
> > > > Nit: int i, resv = 0;
> > > >
> > > > I can make these changes but I suspect this series will go via IOMMU
> > > > tree, let me know how you want to handle it.
> > > >
> > > > Lorenzo
> > > >
> > > >> + node = iort_find_dev_node(dev);
> > > >> + if (!node)
> > > >> + return -ENODEV;
> > > >> +
> > >
> > > I'd suggest we also want a comment here to clarify that we're currently
> > > assuming straightforward topologies where all mappings for a given root
> > > complex/named component target the same ITS group. Otherwise we're
> > going
> > > to need somewhat more logic to iterate the its_node processing over
> > > every mapping (or every alias in the PCI case), but avoid creating
> > > duplicate entries.
> >
> > You have a point and we have time to update the code. Short of reserving
> > all ITS regions for every device that maps to one at least, we could (even
> > pre-compute instead of looking it up on the fly) create a list of ITS
> > identifiers a given IORT node may map to and use that to reserve the
> > regions.
>
> I am trying to understand the use case scenario discussed here.
> Apologies if it is a dumb query.
>
> My understanding is that, it is possible to have a PCI RC iort node
> mapped to multiple ITS group nodes. That is perfectly fine and given
> a dev input RID we can identify the ITS group the device points to
> using - iort_node_map_id().
>
> But the above discussion seems to suggest that there might be
> situations where we have to go through all the mapped ITS groups and
> identify all the ITSs associated with the RC. Clearly I am missing
> something.
I reckon Robin was referring to this:
https://patchwork.kernel.org/patch/9757911/
Does this help ?
Thanks,
Lorenzo
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