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x86/PCI: allocate space from the end of a region, not the beginning

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Allocate from the end of a region, not the beginning.

For example, if we need to allocate 0x800 bytes for a device on bus
0000:00 given these resources:

    [mem 0xbff00000-0xdfffffff] PCI Bus 0000:00
      [mem 0xc0000000-0xdfffffff] PCI Bus 0000:02

the available space at [mem 0xbff00000-0xbfffffff] is passed to the
alignment callback (pcibios_align_resource()).  Prior to this patch, we
would put the new 0x800 byte resource at the beginning of that available
space, i.e., at [mem 0xbff00000-0xbff007ff].

With this patch, we put it at the end, at [mem 0xbffff800-0xbfffffff].

Reference: https://bugzilla.kernel.org/show_bug.cgi?id=16228#c41
Signed-off-by: Bjorn Helgaas <bjorn.helgaas@hp.com>
Signed-off-by: Jesse Barnes <jbarnes@virtuousgeek.org>
This commit is contained in:
Bjorn Helgaas 2010-10-26 15:41:44 -06:00 committed by Jesse Barnes
parent b126b4703a
commit dc9887dc02
1 changed files with 11 additions and 6 deletions
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17
arch/x86/pci/i386.c
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@ -65,16 +65,21 @@ pcibios_align_resource(void *data, const struct resource *res,
resource_size_t size, resource_size_t align)
{
struct pci_dev *dev = data;
resource_size_t start = res->start;
resource_size_t start = round_down(res->end - size + 1, align);
if (res->flags & IORESOURCE_IO) {
if (skip_isa_ioresource_align(dev))
return start;
if (start & 0x300)
start = (start + 0x3ff) & ~0x3ff;
/*
* If we're avoiding ISA aliases, the largest contiguous I/O
* port space is 256 bytes. Clearing bits 9 and 10 preserves
* all 256-byte and smaller alignments, so the result will
* still be correctly aligned.
*/
if (!skip_isa_ioresource_align(dev))
start &= ~0x300;
} else if (res->flags & IORESOURCE_MEM) {
if (start < BIOS_END)
start = BIOS_END;
start = res->end; /* fail; no space */
}
return start;
}