On Mon, Dec 19, 2016 at 04:21:57PM +0100, Michal Hocko wrote:
>On Sun 18-12-16 14:47:50, Wei Yang wrote:
>> 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.
>[...]
>> 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.
>>
>> @@ -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);
>
>
>>
>> --
>> 2.5.0
>>
>> --
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>
>--
>Michal Hocko
>SUSE Labs
--
Wei Yang
Help you, Help me
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