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| author | Ivan Kokshaysky <ink@jurassic.park.msu.ru> | 2005-06-07 04:07:02 +0400 |
|---|---|---|
| committer | Greg Kroah-Hartman <gregkh@suse.de> | 2005-07-01 13:35:50 -0700 |
| commit | 90b54929b626c80056262d9d99b3f48522e404d0 (patch) | |
| tree | d5cb91ff7bd0ac9ffeab5f7bf68235e8b35d050c /drivers/pci | |
| parent | a03fa955576af50df80bec9127b46ef57e0877c0 (diff) | |
| download | linux-90b54929b626c80056262d9d99b3f48522e404d0.tar.gz linux-90b54929b626c80056262d9d99b3f48522e404d0.tar.bz2 linux-90b54929b626c80056262d9d99b3f48522e404d0.zip | |
[PATCH] PCI: handle subtractive decode pci-pci bridge better
With the number of PCI bus resources increased to 8, we can
handle the subtractive decode PCI-PCI bridge like a normal
bridge, taking into account standard PCI-PCI bridge windows
(resources 0-2). This helps to avoid problems with peer-to-peer DMA
behind such bridges, poor performance for MMIO ranges outside bridge
windows and prefetchable vs. non-prefetchable memory issues.
To reflect the fact that such bridges do forward all addresses to
the secondary bus (transparency), remaining bus resources 3-7 are
linked to resources 0-4 of the primary bus. These resources will be
used as fallback by resource management code if allocation from
standard bridge windows fails for some reason.
Signed-off-by: Ivan Kokshaysky <ink@jurassic.park.msu.ru>
Acked-by: Dominik Brodowski <linux@dominikbrodowski.net>
Signed-off-by: Greg Kroah-Hartman <gregkh@suse.de>
Diffstat (limited to 'drivers/pci')
| -rw-r--r-- | drivers/pci/probe.c | 5 |
1 files changed, 2 insertions, 3 deletions
diff --git a/drivers/pci/probe.c b/drivers/pci/probe.c index 9392ff3fb803..df3bdae2040f 100644 --- a/drivers/pci/probe.c +++ b/drivers/pci/probe.c @@ -239,9 +239,8 @@ void __devinit pci_read_bridge_bases(struct pci_bus *child) if (dev->transparent) { printk(KERN_INFO "PCI: Transparent bridge - %s\n", pci_name(dev)); - for(i = 0; i < PCI_BUS_NUM_RESOURCES; i++) - child->resource[i] = child->parent->resource[i]; - return; + for(i = 3; i < PCI_BUS_NUM_RESOURCES; i++) + child->resource[i] = child->parent->resource[i - 3]; } for(i=0; i<3; i++) |