Messages in this thread | | | Date | Wed, 21 Dec 2016 08:51:16 +0100 | From | Michal Hocko <> | Subject | Re: [PATCH V2 2/2] mm/memblock.c: check return value of memblock_reserve() in memblock_virt_alloc_internal() |
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On Tue 20-12-16 16:48:23, Wei Yang wrote: > 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;
So how exactly does the reserve fail when memblock_find_in_range_node found a suitable range for the given size?
> >> > >> 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.
yes, my bad. I have missed this. Sorry about the confusion.
-- Michal Hocko SUSE Labs
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