Messages in this thread | | | Date | Wed, 21 Nov 2018 09:35:59 +0100 | From | Peter Zijlstra <> | Subject | Re: [PATCH 20/25] sched/kcpustat: Introduce vtime-aware kcpustat accessor |
| |
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?
| |