Re: [PATCH tip/core/rcu 08/10] rcu: Add a TINY_PREEMPT_RCU
From: Steven Rostedt
Date: Tue Aug 17 2010 - 10:54:49 EST
On Tue, 2010-08-17 at 10:16 -0400, Mathieu Desnoyers wrote:
> * Steven Rostedt (rostedt@xxxxxxxxxxx) wrote:
> > If we are this concerned, what about just doing:
> >
> > --t->rcu_read_lock_nesting;
> > if (ACCESS_ONCE(t->rcu_read_lock_nesting) == 0 &&
> > unlikely((ACCESS_ONCE(t->rcu_read_unlock_special)))
>
> I'd be concerned by the fact that there is no strong ordering guarantee
> that the non-volatile --t->rcu_read_lock_nesting is done before
> ACCESS_ONCE(t->rcu_read_unlock_special).
>
> My concern is that the compiler might be allowed to turn your code into:
>
> if (ACCESS_ONCE(t->rcu_read_lock_nesting) == 1 &&
> unlikely((ACCESS_ONCE(t->rcu_read_unlock_special))) {
> --t->rcu_read_lock_nesting;
> do_something();
> } else
> --t->rcu_read_lock_nesting;
That just seems to break all sorts of rules.
>
> So whether or not this could be done by the compiler depending on the
> various definitions of volatile, I strongly recommend against using
> volatile accesses to provide compiler ordering guarantees. It is bad in
> terms of code documentation (we don't document _what_ is ordered) and it
> is also bad because the volatile ordering guarantees seems to be
> very easy to misinterpret.
Yes, volatile does not guarantee ordering of other accesses, but it
should at least guarantee ordering of access to the thing that is
volatile.
b++;
a++;
c = ACCESS_ONCE(a);
'b++' can be moved to anywhere. But I'm pretty sure the compiler is not
allowed to move the 'a++' after the ACCESS_ONCE(a) because it is the
thing that is volatile. We are telling the compiler that 'a' can change
outside our scope, which to me is the same as doing:
a++;
c = some_global_function(&a);
Where, the compiler does not know the result of 'a' and can not move the
'a++'.
Maybe I'm wrong, and need to verify this with a compiler expert. But
what's the use of volatile if it can't protect the ordering of what is
volatile from itself.
>
> ACCESS_ONCE() should be only that: a macro that tells the access should
> be performed only once. Why are we suddenly presuming it should have any
> ordering semantic ?
Only ordering with the variable that is volatile. It has no ordering to
any other variable.
>
> It should be totally valid to create arch-specific ACCESS_ONCE() macros
> that only perform the "read once", without the ordering guarantees
> provided by the current ACCESS_ONCE() "volatile" implementation. The
> following code is only for unsigned long, but you get the idea: there is
> no volatile at all, and I ensure that "val" is only read once by using
> the "+m" (val) constraint, telling the compiler (falsely) that the
> assembler is modifying the value (it therefore has a side-effect), so
> gcc won't be tempted to re-issue the assembly statement.
>
> static inline unsigned long arch_access_once(unsigned long val)
> {
> unsigned long ret;
>
> #if (__BITS_PER_LONG == 32)
> asm ("movl %1,%0": "=r" (ret), "+m" (val));
> #else
> asm ("movq %1,%0": "=r" (ret), "+m" (val));
> #endif
> }
Heck, this is too much micro optimization. We could just be safe and do
the:
--t->rcu_read_lock_nesting;
barrier();
if (ACCESS_ONCE(t->rcu_read_lock_nesting) == 0 &&
unlikely((ACCESS_ONCE(t->rcu_read_unlock_special)))
And be done with it.
-- Steve
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