Re: [Xen-devel] [v3][PATCH 07/16] hvmloader/pci: skip reserved ranges

["Chen, Tiejun" <tiejun.chen@xxxxxxxxx>]

On 2015/6/17 16:05, Jan Beulich wrote:
> > > > On 17.06.15 at 09:54, <tiejun.chen@xxxxxxxxx> wrote:
> > On 2015/6/17 15:19, Jan Beulich wrote:
> > > > > > On 17.06.15 at 09:10, <tiejun.chen@xxxxxxxxx> wrote:
> > > > Yeah, this may waste some spaces in this worst case but I this think
> > > > this can guarantee our change don't impact on the original expectation,
> > > > right?
> > > "Some space" may be multiple Gb (e.g. the frame buffer of a graphics
> > Sure.
> >
> > > card), which is totally unacceptable.
> > But then I don't understand what's your way. How can we fit all pci
> > devices just with "the smallest power-of-2 region enclosing the reserved
> > device memory"?
> >
> > For example, the whole pci memory is sitting at
> > [0xa0000000, 0xa2000000]. And there are two PCI devices, A and B. Note
> > each device needs to be allocated with 0x1000000. So if without
> > concerning RMRR,
> >
> > A. [0xa0000000,0xa1000000]
> > B. [0xa1000000,0xa2000000]
> >
> > But if one RMRR resides at [0xa0f00000, 0xa1f00000] which obviously
> > generate its own alignment with 0x1000000. So the pci memory is expended
> > as [0xa0000000, 0xa3000000], right?
> >
> > Then actually the whole pci memory can be separated three segments like,
> >
> > #1. [0xa0000000, 0xa0f00000]
> > #2. [0xa0f00000, 0xa1f00000] -> RMRR would occupy
> > #3. [0xa1f00000, 0xa3000000]
> >
> > So just #3 can suffice to allocate but just for one device, right?
> Right, i.e. this isn't even sufficient - you need [a0000000,a3ffffff]
> to fit everything (but of course you can put smaller BARs into the
> unused ranges [a0000000,a0efffff] and [a1f00000,a1ffffff]).
Yes, I knew there's this sort of hole that we should use efficiently as 
you said. And I also thought about this way previously but current pci 
allocation framework isn't friend to implement this easily,
    /* Assign iomem and ioport resources in descending order of size. */
    for ( i = 0; i < nr_bars; i++ )
    {

I mean it isn't easy to calculate what's the most sufficient size in 
advance, and its also difficult to find to fit a appropriate pci device 
into those "holes", so see below,
> That's why I said it's not going to be tricky to get all corner cases
> right _and_ not use up more space than needed.
>
> > > > > ought to work out the smallest power-of-2 region enclosing the
> > > > Okay. I remember the smallest size of a given PCI I/O space is 8 bytes,
> > > > and the smallest size of a PCI memory space is 16 bytes. So
> > > >
> > > > /* At least 16 bytes to align a PCI BAR size. */
> > > > uint64_t align = 16;
> > > >
> > > > reserved_start = memory_map.map[j].addr;
> > > > reserved_size = memory_map.map[j].size;
> > > >
> > > > reserved_start = (reserved_star + align) & ~(align - 1);
> > > > reserved_size = (reserved_size + align) & ~(align - 1);
> > > >
> > > > Is this correct?
> > > Simply aligning the region doesn't help afaict. You need to fit it
> > > with the other MMIO allocations.
> > I guess you're saying just those mmio allocations conflicting with RMRR?
> > But we don't know these exact addresses until we finalize to allocate
> > them, right?
> That's the point - you need to allocate them _around_ the reserved
> regions.
Something hits me to generate another idea,

#1. Still allocate all devices as before.
#2. Lookup all actual bars to check if they're conflicting RMRR

We can skip these bars to keep zero. Then later it would make lookup easily.

#3. Need to reallocate these conflicting bars.
#3.1 Trying to reallocate them with the remaining resources
#3.2 If the remaining resources aren't enough, we need to allocate them 
from high_mem_resource.
I just feel this way may be easy and better. And even, this way also can 
help terminate the preexisting allocation failures, right?
Thanks
Tiejun

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