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ARM: LMB: Convert arm_memory_present() to use LMB memory information 

Signed-off-by: Russell King <rmk+kernel@arm.linux.org.uk>
This commit is contained in:
Russell King 2010-07-01 12:00:57 +01:00
parent 98864ff58d
commit eda2e5dcc9
1 changed files with 6 additions and 7 deletions
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13
arch/arm/mm/init.c
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@ -257,17 +257,16 @@ int pfn_valid(unsigned long pfn)
}
EXPORT_SYMBOL(pfn_valid);
static void arm_memory_present(struct meminfo *mi)
static void arm_memory_present(void)
{
}
#else
static void arm_memory_present(struct meminfo *mi)
static void arm_memory_present(void)
{
int i;
for_each_bank(i, mi) {
struct membank *bank = &mi->bank[i];
memory_present(0, bank_pfn_start(bank), bank_pfn_end(bank));
}
for (i = 0; i < memblock.memory.cnt; i++)
memory_present(0, memblock_start_pfn(&memblock.memory, i),
memblock_end_pfn(&memblock.memory, i));
}
#endif
@ -320,7 +319,7 @@ void __init bootmem_init(void)
* Sparsemem tries to allocate bootmem in memory_present(),
* so must be done after the fixed reservations
*/
arm_memory_present(mi);
arm_memory_present();
/*
* sparse_init() needs the bootmem allocator up and running.