On Mon, Dec 19, 2016 at 04:21:57PM +0100, Michal Hocko wrote:

On Sun 18-12-16 14:47:50, Wei Yang wrote:

quoted

memblock_reserve() may fail in case there is not enough regions.

Have you seen this happenning in the real setups or this is a by-review
driven change?

This is a by-review driven change.

[...]

quoted

 again:
 	alloc = memblock_find_in_range_node(size, align, min_addr, max_addr,
 					    nid, flags);
-	if (alloc)
+	if (alloc && !memblock_reserve(alloc, size))
 		goto done;
 
 	if (nid != NUMA_NO_NODE) {
 		alloc = memblock_find_in_range_node(size, align, min_addr,
 						    max_addr, NUMA_NO_NODE,
 						    flags);
-		if (alloc)
+		if (alloc && !memblock_reserve(alloc, size))
 			goto done;
 	}

This doesn't look right. You can end up leaking the first allocated
range.

Hmm... why?

If first memblock_reserve() succeed, it will jump to done, so that no 2nd
allocation.
If the second executes, it means the first allocation failed.
memblock_find_in_range_node() doesn't modify the memblock, it just tell you
there is a proper memory region available.

quoted

@@ -1303,7 +1302,6 @@ static void * __init memblock_virt_alloc_internal(
 
 	return NULL;
 done:
-	memblock_reserve(alloc, size);
 	ptr = phys_to_virt(alloc);
 	memset(ptr, 0, size);

quoted

 
-- 
2.5.0

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-- 
Michal Hocko
SUSE Labs

-- 
Wei Yang
Help you, Help me

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