linux-kernel - Re: SLUB: sysfs lets root force slab order below required minimum, causing memory corruption

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Date:   Tue, 3 Mar 2020 17:26:14 -0800 (PST)
From:   David Rientjes <rientjes@...gle.com>
To:     Jann Horn <jannh@...gle.com>
cc:     Christoph Lameter <cl@...ux.com>,
        Pekka Enberg <penberg@...nel.org>,
        Joonsoo Kim <iamjoonsoo.kim@....com>,
        Andrew Morton <akpm@...ux-foundation.org>,
        Linux-MM <linux-mm@...ck.org>,
        kernel list <linux-kernel@...r.kernel.org>,
        Kees Cook <keescook@...omium.org>,
        Matthew Garrett <mjg59@...gle.com>
Subject: Re: SLUB: sysfs lets root force slab order below required minimum,
 causing memory corruption

On Wed, 4 Mar 2020, Jann Horn wrote:

> Hi!
> 
> FYI, I noticed that if you do something like the following as root,
> the system blows up pretty quickly with error messages about stuff
> like corrupt freelist pointers because SLUB actually allows root to
> force a page order that is smaller than what is required to store a
> single object:
> 
>     echo 0 > /sys/kernel/slab/task_struct/order
> 
> The other SLUB debugging options, like red_zone, also look kind of
> suspicious with regards to races (either racing with other writes to
> the SLUB debugging options, or with object allocations).
> 

Thanks for the report, Jann.  To address the most immediate issue, 
allowing a smaller order than allowed, I think we'd need something like 
this.

I can propose it as a formal patch if nobody has any alternate 
suggestions?
---
 mm/slub.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/mm/slub.c b/mm/slub.c
--- a/mm/slub.c
+++ b/mm/slub.c
@@ -3598,7 +3598,7 @@ static int calculate_sizes(struct kmem_cache *s, int forced_order)
 	 */
 	size = ALIGN(size, s->align);
 	s->size = size;
-	if (forced_order >= 0)
+	if (forced_order >= slab_order(size, 1, MAX_ORDER, 1))
 		order = forced_order;
 	else
 		order = calculate_order(size);