linux-kernel - Memory allocator semantics

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Message-ID: <20140102203320.GA27615@linux.vnet.ibm.com>
Date:	Thu, 2 Jan 2014 12:33:20 -0800
From:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
To:	linux-mm@...ck.org, linux-kernel@...r.kernel.org
Cc:	cl@...ux-foundation.org, penberg@...nel.org, mpm@...enic.com
Subject: Memory allocator semantics

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."

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