Memory allocator semantics

Paul E. McKenney
2014-01-02T20:33:41Z

Hello!

>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?

Thanx, Paul

PS. To the question "Why would anyone care about (A)?", then answer
is "Inquiring programming-language memory-model designers want
to know."