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authorIvan Kokshaysky <ink@jurassic.park.msu.ru>2005-06-06 20:07:02 -0400
committerGreg Kroah-Hartman <gregkh@suse.de>2005-07-01 16:35:50 -0400
commit90b54929b626c80056262d9d99b3f48522e404d0 (patch)
treed5cb91ff7bd0ac9ffeab5f7bf68235e8b35d050c /drivers/pci/probe.c
parenta03fa955576af50df80bec9127b46ef57e0877c0 (diff)
[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/probe.c')
-rw-r--r--drivers/pci/probe.c5
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)
239 239
240 if (dev->transparent) { 240 if (dev->transparent) {
241 printk(KERN_INFO "PCI: Transparent bridge - %s\n", pci_name(dev)); 241 printk(KERN_INFO "PCI: Transparent bridge - %s\n", pci_name(dev));
242 for(i = 0; i < PCI_BUS_NUM_RESOURCES; i++) 242 for(i = 3; i < PCI_BUS_NUM_RESOURCES; i++)
243 child->resource[i] = child->parent->resource[i]; 243 child->resource[i] = child->parent->resource[i - 3];
244 return;
245 } 244 }
246 245
247 for(i=0; i<3; i++) 246 for(i=0; i<3; i++)