Re: bit fields && data tearing

H. Peter Anvin
2014-09-07T23:39:57Z

How many PARISC systems do we have that actually do real work on Linux?

On September 7, 2014 4:36:55 PM PDT, "Paul E. McKenney" <paulmck@linux.vnet.ibm.com> wrote:
>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"
><paulmck@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
>>
>> --
>> Sent from my mobile phone. Please pardon brevity and lack of
>formatting.
>>

--
Sent from my mobile phone. Please pardon brevity and lack of formatting.