Re: [PATCH V2 2/2] mm/memblock.c: check return value of memblock_reserve() in memblock_virt_alloc_internal()

From: Michal Hocko
Date: Wed Dec 21 2016 - 02:51:24 EST


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