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PCI: Make minimum bridge window alignment reference more obvious

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Calculations related to bridge window size contain literal 20 that is the
minimum alignment for a bridge window. Make the code more obvious by
converting the literal 20 to __ffs(SZ_1M).

Link: https://lore.kernel.org/r/20240507102523.57320-8-ilpo.jarvinen@linux.intel.com
Signed-off-by: Ilpo Järvinen <ilpo.jarvinen@linux.intel.com>
[bhelgaas: squash https://lore.kernel.org/r/20240612093250.17544-1-ilpo.jarvinen@linux.intel.com]
Signed-off-by: Bjorn Helgaas <bhelgaas@google.com>
Reviewed-by: Mika Westerberg <mika.westerberg@linux.intel.com>
This commit is contained in:
Ilpo Järvinen 2024-05-07 13:25:22 +03:00 committed by Bjorn Helgaas
parent 903534fa7d
commit 8fa0a44eb0
1 changed files with 4 additions and 2 deletions
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6
drivers/pci/setup-bus.c
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@ -14,6 +14,7 @@
* tighter packing. Prefetchable range support.
*/
#include <linux/bitops.h>
#include <linux/init.h>
#include <linux/kernel.h>
#include <linux/module.h>
@ -21,6 +22,7 @@
#include <linux/errno.h>
#include <linux/ioport.h>
#include <linux/cache.h>
#include <linux/sizes.h>
#include <linux/slab.h>
#include <linux/acpi.h>
#include "pci.h"
@ -957,7 +959,7 @@ static inline resource_size_t calculate_mem_align(resource_size_t *aligns,
for (order = 0; order <= max_order; order++) {
resource_size_t align1 = 1;
align1 <<= (order + 20);
align1 <<= order + __ffs(SZ_1M);
if (!align)
min_align = align1;
@ -1047,7 +1049,7 @@ static int pbus_size_mem(struct pci_bus *bus, unsigned long mask,
* resources.
*/
align = pci_resource_alignment(dev, r);
order = __ffs(align) - 20;
order = __ffs(align) - __ffs(SZ_1M);
if (order < 0)
order = 0;
if (order >= ARRAY_SIZE(aligns)) {