[PATCH] ARM: use memblock memory regions for "System RAM" I/O resources
Russell King - ARM Linux
linux at arm.linux.org.uk
Wed Jan 12 18:52:23 EST 2011
On Wed, Jan 12, 2011 at 03:22:40PM -0800, Dima Zavin wrote:
> @@ -518,8 +518,9 @@ setup_ramdisk(int doload, int prompt, int image_start, unsigned int rd_sz)
> }
>
> static void __init
> -request_standard_resources(struct meminfo *mi, struct machine_desc *mdesc)
> +request_standard_resources(struct machine_desc *mdesc)
Almost there. This line can now be combined with the preceding line
without wrapping over the 80 column limit.
> {
> + struct memblock_type *mem = &memblock.memory;
> struct resource *res;
> int i;
>
> @@ -528,14 +529,11 @@ request_standard_resources(struct meminfo *mi, struct machine_desc *mdesc)
> kernel_data.start = virt_to_phys(_sdata);
> kernel_data.end = virt_to_phys(_end - 1);
>
> - for (i = 0; i < mi->nr_banks; i++) {
> - if (mi->bank[i].size == 0)
> - continue;
> -
> + for (i = 0; i < mem->cnt; i++) {
Hmm, probably shouldn't rely on that - we should be using the accessors
which memblock people have provided for us.
for_each_memblock(memory, mem) {
> res = alloc_bootmem_low(sizeof(*res));
> res->name = "System RAM";
> - res->start = mi->bank[i].start;
> - res->end = mi->bank[i].start + mi->bank[i].size - 1;
> + res->start = mem->regions[i].base;
> + res->end = mem->regions[i].base + mem->regions[i].size - 1;
res->start = __pfn_to_phys(memblock_region_memory_base_pfn(mem));
res->end = __pfn_to_phys(memblock_region_memory_end_pfn(mem)) - 1;
And I'd say ignore the 80-column limit on that because of the excessively
long memblock function names.
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