multiple separate pci bridges ...
Rob Baxter
robb at synergymicro.com
Wed Jan 7 02:33:05 EST 2004
On Tue, Jan 06, 2004 at 03:45:45PM +0100, Sven Luther wrote:
>
> On Tue, Jan 06, 2004 at 03:40:30PM +0100, Geert Uytterhoeven wrote:
> > On Tue, 6 Jan 2004, Sven Luther wrote:
> > > On Tue, Jan 06, 2004 at 07:00:24PM +1100, Benjamin Herrenschmidt wrote:
> > > > Type 0 is an access to the primary segment (doesn't contain a bus
> > > > number), type 1 is to be forwarded to another bus segment by a P2P
> > > > bridge. So for anything directly attached to the host bridge, it's a
> > > > type 0 access. Anything else is type 1. Typically, if the bus number of
> > > > your "target" == hose->first_busno, it's type 0, else type 1
> > >
> > > Yep, except we have two pci controllers, and it should be type 0 for
> > > both of them.
> >
> > Yep, two hoses, with different hose->first_busno.
>
> A, yes, right.
>
> > > > They could have appeared as on-chip PCI devices on a "pseudo-bus", but
> > > > we can eventually just match with the host's PCI device.
> > >
> > > Ok. but this can also be faked or something ? But, how can we match with
> > > the host PCI device, if we are going to hide it ?
> >
> > Hide the memory BAR only, not the full PCI device.
>
> Ok, will try that.
We hide all of the PCI resources (i.e., PCI BARs) by initializing all of
resources to zero. This allows us to load drivers (e.g., ethernet) specific
to the Discovery. Registers internal to the Discovery (via the PowerPC
bus) need to be initialized and queried for memory, PCI I/O, PCI memory,
and PCI prefecthable memory mappings (hopefully your firmware does this).
Here's the code from our pcibios_fixup:
dev = NULL;
while ((dev = pci_find_device(PCI_VENDOR_ID_GALILEO,
PCI_DEVICE_ID_GALILEO_GT64260, dev))) {
for (i = 0; i < DEVICE_COUNT_RESOURCE; i++) {
dev->resource[i].flags = 0;
dev->resource[i].start = 0;
dev->resource[i].end = 0;
}
}
>
> BTW, is there any reason the L2 cache is disabled by default in the
> 2.4.x kernels ?
We have it initialized and enabled.
>
> Friendly,
>
> Sven Luther
>
--
-Rob Baxter
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