linux-kernel - Re: Memory allocator semantics

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Message-ID: <52F60699.8010204@iki.fi>
Date:	Sat, 08 Feb 2014 12:27:37 +0200
From:	Pekka Enberg <penberg@....fi>
To:	paulmck@...ux.vnet.ibm.com, linux-mm@...ck.org,
	linux-kernel@...r.kernel.org
CC:	cl@...ux-foundation.org, penberg@...nel.org, mpm@...enic.com
Subject: Re: Memory allocator semantics

Hi Paul,

On 01/02/2014 10:33 PM, Paul E. McKenney wrote:
>  From what I can see, the Linux-kernel's SLAB, SLOB, and SLUB memory
> allocators would deal with the following sort of race:
>
> A.	CPU 0: r1 = kmalloc(...); ACCESS_ONCE(gp) = r1;
>
> 	CPU 1: r2 = ACCESS_ONCE(gp); if (r2) kfree(r2);
>
> However, my guess is that this should be considered an accident of the
> current implementation rather than a feature.  The reason for this is
> that I cannot see how you would usefully do (A) above without also allowing
> (B) and (C) below, both of which look to me to be quite destructive:
>
> B.	CPU 0: r1 = kmalloc(...);  ACCESS_ONCE(shared_x) = r1;
>
>          CPU 1: r2 = ACCESS_ONCE(shared_x); if (r2) kfree(r2);
>
> 	CPU 2: r3 = ACCESS_ONCE(shared_x); if (r3) kfree(r3);
>
> 	This results in the memory being on two different freelists.
>
> C.      CPU 0: r1 = kmalloc(...);  ACCESS_ONCE(shared_x) = r1;
>
> 	CPU 1: r2 = ACCESS_ONCE(shared_x); r2->a = 1; r2->b = 2;
>
> 	CPU 2: r3 = ACCESS_ONCE(shared_x); if (r3) kfree(r3);
>
> 	CPU 3: r4 = kmalloc(...);  r4->s = 3; r4->t = 4;
>
> 	This results in the memory being used by two different CPUs,
> 	each of which believe that they have sole access.
>
> But I thought I should ask the experts.
>
> So, am I correct that kernel hackers are required to avoid "drive-by"
> kfree()s of kmalloc()ed memory?

So to be completely honest, I don't understand what is the race in (A) 
that concerns the *memory allocator*.  I also don't what the memory 
allocator can do in (B) and (C) which look like double-free and 
use-after-free, respectively, to me. :-)

                       Pekka
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