On 8/15/07, Christoph Lameter <[email protected]> wrote:
> On Tue, 14 Aug 2007, Lucio Correia wrote:
>
> > > SLAB boots because it falls back to node 0 for the control structures. So
> > > it creates useless control structures for node 1. These are then never
> > > used since any allocation attempt to node 1 falls back to node 0.
> >
> > Hi Christoph,
> >
> > Shouldn't SLUB falls back to other node also for the case it can't
> > allocate memory?
>
> Yes SLUB will fall back but not during bootstrap. Bootstrap needs to
> carefully place structures on the right nodes. We fail during bootstrap
> because there is *no* memory available on it.
Sure, you want to have the structures on the right node if possible.
But seeing as there's no memory available, what is wrong with just
falling back?
Something like:
diff --git a/mm/slub.c b/mm/slub.c
index 9b2d617..0fc29d2 100644
--- a/mm/slub.c
+++ b/mm/slub.c
@@ -1876,6 +1876,8 @@ static struct kmem_cache_node * __init early_kmem_cache_no
BUG_ON(kmalloc_caches->size < sizeof(struct kmem_cache_node));
page = new_slab(kmalloc_caches, gfpflags | GFP_THISNODE, node);
+ if (!page)
+ page = new_slab(kmalloc_caches, gfpflags, node);
BUG_ON(!page);
n = page->freelist;
-
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