bit fields && data tearing

[H. Peter Anvin]

I'm confused why storing 0x0102 would be a problem.  I think gcc does that even on other cpus.

More atomicity can't hurt, can it?

On September 7, 2014 4:00:19 PM PDT, "Paul E. McKenney" <paulmck at linux.vnet.ibm.com> wrote:
>On Sun, Sep 07, 2014 at 12:04:47PM -0700, James Bottomley wrote:
>> On Sun, 2014-09-07 at 09:21 -0700, Paul E. McKenney wrote:
>> > On Sat, Sep 06, 2014 at 10:07:22PM -0700, James Bottomley wrote:
>> > > On Thu, 2014-09-04 at 21:06 -0700, Paul E. McKenney wrote:
>> > > > On Thu, Sep 04, 2014 at 10:47:24PM -0400, Peter Hurley wrote:
>> > > > > Hi James,
>> > > > > 
>> > > > > On 09/04/2014 10:11 PM, James Bottomley wrote:
>> > > > > > On Thu, 2014-09-04 at 17:17 -0700, Paul E. McKenney wrote:
>> > > > > >> +And there are anti-guarantees:
>> > > > > >> +
>> > > > > >> + (*) These guarantees do not apply to bitfields, because
>compilers often
>> > > > > >> +     generate code to modify these using non-atomic
>read-modify-write
>> > > > > >> +     sequences.  Do not attempt to use bitfields to
>synchronize parallel
>> > > > > >> +     algorithms.
>> > > > > >> +
>> > > > > >> + (*) Even in cases where bitfields are protected by
>locks, all fields
>> > > > > >> +     in a given bitfield must be protected by one lock. 
>If two fields
>> > > > > >> +     in a given bitfield are protected by different
>locks, the compiler's
>> > > > > >> +     non-atomic read-modify-write sequences can cause an
>update to one
>> > > > > >> +     field to corrupt the value of an adjacent field.
>> > > > > >> +
>> > > > > >> + (*) These guarantees apply only to properly aligned and
>sized scalar
>> > > > > >> +     variables.  "Properly sized" currently means "int"
>and "long",
>> > > > > >> +     because some CPU families do not support loads and
>stores of
>> > > > > >> +     other sizes.  ("Some CPU families" is currently
>believed to
>> > > > > >> +     be only Alpha 21064.  If this is actually the case,
>a different
>> > > > > >> +     non-guarantee is likely to be formulated.)
>> > > > > > 
>> > > > > > This is a bit unclear.  Presumably you're talking about
>definiteness of
>> > > > > > the outcome (as in what's seen after multiple stores to the
>same
>> > > > > > variable).
>> > > > > 
>> > > > > No, the last conditions refers to adjacent byte stores from
>different
>> > > > > cpu contexts (either interrupt or SMP).
>> > > > > 
>> > > > > > The guarantees are only for natural width on Parisc as
>well,
>> > > > > > so you would get a mess if you did byte stores to adjacent
>memory
>> > > > > > locations.
>> > > > > 
>> > > > > For a simple test like:
>> > > > > 
>> > > > > struct x {
>> > > > > 	long a;
>> > > > > 	char b;
>> > > > > 	char c;
>> > > > > 	char d;
>> > > > > 	char e;
>> > > > > };
>> > > > > 
>> > > > > void store_bc(struct x *p) {
>> > > > > 	p->b = 1;
>> > > > > 	p->c = 2;
>> > > > > }
>> > > > > 
>> > > > > on parisc, gcc generates separate byte stores
>> > > > > 
>> > > > > void store_bc(struct x *p) {
>> > > > >    0:	34 1c 00 02 	ldi 1,ret0
>> > > > >    4:	0f 5c 12 08 	stb ret0,4(r26)
>> > > > >    8:	34 1c 00 04 	ldi 2,ret0
>> > > > >    c:	e8 40 c0 00 	bv r0(rp)
>> > > > >   10:	0f 5c 12 0a 	stb ret0,5(r26)
>> > > > > 
>> > > > > which appears to confirm that on parisc adjacent byte data
>> > > > > is safe from corruption by concurrent cpu updates; that is,
>> > > > > 
>> > > > > CPU 0                | CPU 1
>> > > > >                      |
>> > > > > p->b = 1             | p->c = 2
>> > > > >                      |
>> > > > > 
>> > > > > will result in p->b == 1 && p->c == 2 (assume both values
>> > > > > were 0 before the call to store_bc()).
>> > > > 
>> > > > What Peter said.  I would ask for suggestions for better
>wording, but
>> > > > I would much rather be able to say that single-byte reads and
>writes
>> > > > are atomic and that aligned-short reads and writes are also
>atomic.
>> > > > 
>> > > > Thus far, it looks like we lose only very old Alpha systems, so
>unless
>> > > > I hear otherwise, I update my patch to outlaw these very old
>systems.
>> > > 
>> > > This isn't universally true according to the architecture manual.
> The
>> > > PARISC CPU can make byte to long word stores atomic against the
>memory
>> > > bus but not against the I/O bus for instance.  Atomicity is a
>property
>> > > of the underlying substrate, not of the CPU.  Implying that
>atomicity is
>> > > a CPU property is incorrect.
>> > 
>> > OK, fair point.
>> > 
>> > But are there in-use-for-Linux PARISC memory fabrics (for normal
>memory,
>> > not I/O) that do not support single-byte and double-byte stores?
>> 
>> For aligned access, I believe that's always the case for the memory
>bus
>> (on both 32 and 64 bit systems).  However, it only applies to machine
>> instruction loads and stores of the same width..  If you mix the
>widths
>> on the loads and stores, all bets are off.  That means you have to
>> beware of the gcc penchant for coalescing loads and stores: if it
>sees
>> two adjacent byte stores it can coalesce them into a short store
>> instead ... that screws up the atomicity guarantees.
>
>OK, that means that to make PARISC work reliably, we need to use
>ACCESS_ONCE() for loads and stores that could have racing accesses.
>If I understand correctly, this will -not- be needed for code guarded
>by locks, even with Peter's examples.
>
>So if we have something like this:
>
>	struct foo {
>		char a;
>		char b;
>	};
>	struct foo *fp;
>
>then this code would be bad:
>
>	fp->a = 1;
>	fp->b = 2;
>
>The reason is (as you say) that GCC would be happy to store 0x0102
>(or vice versa, depending on endianness) to the pair.  We instead
>need:
>
>	ACCESS_ONCE(fp->a) = 1;
>	ACCESS_ONCE(fp->b) = 2;
>
>However, if the code is protected by locks, no problem:
>
>	struct foo {
>		spinlock_t lock_a;
>		spinlock_t lock_b;
>		char a;
>		char b;
>	};
>
>Then it is OK to do the following:
>
>	spin_lock(fp->lock_a);
>	fp->a = 1;
>	spin_unlock(fp->lock_a);
>	spin_lock(fp->lock_b);
>	fp->b = 1;
>	spin_unlock(fp->lock_b);
>
>Or even this, assuming ->lock_a precedes ->lock_b in the locking
>hierarchy:
>
>	spin_lock(fp->lock_a);
>	spin_lock(fp->lock_b);
>	fp->a = 1;
>	fp->b = 1;
>	spin_unlock(fp->lock_a);
>	spin_unlock(fp->lock_b);
>
>Here gcc might merge the assignments to fp->a and fp->b, but that is OK
>because both locks are held, presumably preventing other assignments or
>references to fp->a and fp->b.
>
>On the other hand, if either fp->a or fp->b are referenced outside of
>their
>respective locks, even once, then this last code fragment would still
>need
>ACCESS_ONCE() as follows:
>
>	spin_lock(fp->lock_a);
>	spin_lock(fp->lock_b);
>	ACCESS_ONCE(fp->a) = 1;
>	ACCESS_ONCE(fp->b) = 1;
>	spin_unlock(fp->lock_a);
>	spin_unlock(fp->lock_b);
>
>Does that cover it?  If so, I will update memory-barriers.txt
>accordingly.
>
>							Thanx, Paul

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