* Christoph Lameter (clameter@sgi.com) wrote:
 
quoted text
> > The cpu_read acts as a safeguard checking that we do not change CPU
> > between the read and the cmpxchg. If we are preempted between the "c"
> > read and the cpu_read, we could do a !cpu_node_match(c, node) check that
> > would apply to the wrong cpu.
> 
> C is not pointing to a specific cpu. It can only be used in CPU_xx ops to
> address the currrent cpu.
> 
> > > @@ -1800,19 +1792,19 @@ static void __always_inline slab_free(st
> > >  		 * then any change of cpu_slab will cause the cmpxchg to fail
> > >  		 * since the freelist pointers are unique per slab.
> > >  		 */
> > > -		if (unlikely(page != c->page || c->node < 0)) {
> > > -			__slab_free(s, page, x, addr, c->offset);
> > > +		if (unlikely(page != __CPU_READ(c->page) ||
> > > +					__CPU_READ(c->node) < 0)) {
> > > +			__slab_free(s, page, x, addr, __CPU_READ(c->offset));
> > 
> > And same question as above : what happens if we fail after executing the
> > __slab_free.. is it valid to do it twice ?
> 
> __slab_free is always successful and will never cause a repeat of the 
> loop.

Then what happens if we are migrated between the end of __slab_free and
the CPU_CMPXCHG ? The cmpxchg would fail, and the __slab_free will be
done twice ?

-- 
Mathieu Desnoyers
Computer Engineering Ph.D. Student, Ecole Polytechnique de Montreal
OpenPGP key fingerprint: 8CD5 52C3 8E3C 4140 715F  BA06 3F25 A8FE 3BAE 9A68
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