On Wed, Nov 21, 2018 at 09:18:19AM +0100, Peter Zijlstra wrote:
> On Tue, Nov 20, 2018 at 11:40:22PM +0100, Frederic Weisbecker wrote:
> > On Tue, Nov 20, 2018 at 03:23:06PM +0100, Peter Zijlstra wrote:
> > > On Wed, Nov 14, 2018 at 03:46:04AM +0100, Frederic Weisbecker wrote:
> > >
> > > > +void kcpustat_cputime(struct kernel_cpustat *kcpustat, int cpu,
> > > > + u64 *user, u64 *nice, u64 *system,
> > > > + u64 *guest, u64 *guest_nice)
> > > > +{
> > > > + struct task_struct *curr;
> > > > + struct vtime *vtime;
> > > > + int err;
> > > > +
> > > > + if (!vtime_accounting_enabled()) {
> > > > + kcpustat_cputime_raw(kcpustat, user, nice,
> > > > + system, guest, guest_nice);
> > > > + return;
> > > > + }
> > > > +
> > > > + rcu_read_lock();
> > > > +
> > > > + do {
> > > > + curr = rcu_dereference(kcpustat->curr);
> > >
> > > Like I explained earlier; I don't think the above is correct.
> > > task_struct is itself not RCU protected.
> >
> > But there is at least one put_task_struct() that is enqueued as an RCU
> > callback
> > on release_task(). That patchset (try to) make sure that kcpustat->curr
> > can't
> > be assigned beyond that point.
> >
> > Or did I misunderstand something?
>
> Yeah; release_task() is not the normal exit path. Oleg can probably
> remember how all that works, because I always get lost there :-/
>
> In any case, have a look at task_rcu_dereference(), but that still does
> not explain the rcu_assign_pointer() stuff you use to set
> kcpustat->curr.
Also, why do you need kcpustat->curr at all, the above function has
@cpu, so you can equally use cpu_curr(cpu), no?
