efi: Enforce minimum alignment of 1 page on allocations.

The efi_high_alloc() and efi_low_alloc() functions
use the EFI_ALLOCATE_ADDRESS option to the EFI
function allocate_pages(), which requires a minimum
of page alignment, and rejects all other requests.
The existing code could fail to allocate depending
on allocation size, as although repeated allocation
attempts were made, none were guaranteed to be page
aligned.

Signed-off-by: Roy Franz <roy.franz@linaro.org>
Acked-by: Mark Salter <msalter@redhat.com>
Reviewed-by: Grant Likely <grant.likely@linaro.org>
Signed-off-by: Matt Fleming <matt.fleming@intel.com>
This commit is contained in:
Roy Franz 2013-09-22 15:45:30 -07:00 коммит произвёл Matt Fleming
Родитель 40e4530a00
Коммит 38dd9c02c3
1 изменённых файлов: 16 добавлений и 0 удалений

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@ -101,6 +101,14 @@ static efi_status_t efi_high_alloc(efi_system_table_t *sys_table_arg,
if (status != EFI_SUCCESS)
goto fail;
/*
* Enforce minimum alignment that EFI requires when requesting
* a specific address. We are doing page-based allocations,
* so we must be aligned to a page.
*/
if (align < EFI_PAGE_SIZE)
align = EFI_PAGE_SIZE;
nr_pages = round_up(size, EFI_PAGE_SIZE) / EFI_PAGE_SIZE;
again:
for (i = 0; i < map_size / desc_size; i++) {
@ -179,6 +187,14 @@ static efi_status_t efi_low_alloc(efi_system_table_t *sys_table_arg,
if (status != EFI_SUCCESS)
goto fail;
/*
* Enforce minimum alignment that EFI requires when requesting
* a specific address. We are doing page-based allocations,
* so we must be aligned to a page.
*/
if (align < EFI_PAGE_SIZE)
align = EFI_PAGE_SIZE;
nr_pages = round_up(size, EFI_PAGE_SIZE) / EFI_PAGE_SIZE;
for (i = 0; i < map_size / desc_size; i++) {
efi_memory_desc_t *desc;