[PATCH 07/11] x86, memblock: Set lowest limit for memblock_alloc_base_nid().

[Tang Chen]

memblock_alloc_base_nid() is a common API of memblock. And it calls
memblock_find_in_range_node() with %start = 0, which means it has no
limit for the lowest address by default.

	memblock_find_in_range_node(0, max_addr, size, align, nid);

Since we introduced current_limit_low to memblock, if we have no limit
for the lowest address or we are not sure, we should pass
MEMBLOCK_ALLOC_ACCESSIBLE to %start so that it will be limited by the
default low limit.

dma_contiguous_reserve() and setup_log_buf() will eventually call
memblock_alloc_base_nid() to allocate memory. So if the allocation order
is from low to high, they will allocate memory from the lowest limit
to higher memory.

Signed-off-by: Tang Chen <tangchen@xxxxxxxxxxxxxx>
Reviewed-by: Zhang Yanfei <zhangyanfei@xxxxxxxxxxxxxx>
---
 mm/memblock.c |    3 ++-
 1 files changed, 2 insertions(+), 1 deletions(-)

diff --git a/mm/memblock.c b/mm/memblock.c
index 961d4a5..be8c4d1 100644
--- a/mm/memblock.c
+++ b/mm/memblock.c
@@ -851,7 +851,8 @@ static phys_addr_t __init memblock_alloc_base_nid(phys_addr_t size,
 	/* align @size to avoid excessive fragmentation on reserved array */
 	size = round_up(size, align);
 
-	found = memblock_find_in_range_node(0, max_addr, size, align, nid);
+	found = memblock_find_in_range_node(MEMBLOCK_ALLOC_ACCESSIBLE,
+					    max_addr, size, align, nid);
 	if (found && !memblock_reserve(found, size))
 		return found;
 
-- 
1.7.1

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