Messages in this thread |  | | Date | Sat, 08 Feb 2014 12:27:37 +0200 | From | Pekka Enberg <> | Subject | Re: Memory allocator semantics |
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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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