Subject: Re: bit fields && data tearing

From: "Paul E. McKenney" <[email-masked]>
Date: Sun, 7 Sep 2014 16:36:55 -0700

On Sun, Sep 07, 2014 at 04:17:30PM -0700, H. Peter Anvin wrote:
> 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?

I must defer to James for any additional details on why PARISC systems
don't provide atomicity for partially overlapping stores.  ;-)

							Thanx, Paul

> On September 7, 2014 4:00:19 PM PDT, "Paul E. McKenney" <[email-masked]> 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
> 
> -- 
> Sent from my mobile phone.  Please pardon brevity and lack of formatting.
>