Messages in this thread | | | Subject | Re: [PATCH 2/3] ftrace: use struct pid | From | Dave Hansen <> | Date | Thu, 04 Dec 2008 07:41:31 -0800 |
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On Thu, 2008-12-04 at 05:40 -0800, Eric W. Biederman wrote: > Dave Hansen <dave@linux.vnet.ibm.com> writes: > > On Thu, 2008-12-04 at 04:56 -0800, Dave Hansen wrote: > >> On Thu, 2008-12-04 at 04:42 -0800, Eric W. Biederman wrote: > >> > > >> > > +static void clear_ftrace_pid_task(struct pid **pid) > >> > > +{ > >> > > + struct task_struct *p; > >> > > + > >> > rcu_read_lock(); > >> > > >> > > + do_each_pid_task(*pid, PIDTYPE_PID, p) { > >> > > + clear_tsk_trace_trace(p); > >> > > + } while_each_pid_task(*pid, PIDTYPE_PID, p); > >> > rcu_read_unlock() > >> > > >> > > + put_pid(*pid); > >> > > + > >> > > + *pid = NULL; > >> > > +} > >> > >> Could we get away with sticking the rcu_read_{un}lock() inside > those > >> macros? Those are going to get used in pretty high level code and > we're > >> allowed to nest rcu_read_lock(). No danger of deadlocks or lock > >> inversions. > > > > Why don't any of the other users of do_each_pid_task() use > > rcu_read_lock()? They all seem to be under > read_lock(&tasklist_lock) > > (except one is under a write lock of the same). > > We probably should. Historically read_lock(&tasklist_lock) implies > rcu_read_lock(). And the tasklist lock is what we hold when it is > safe.
So, Dipankar tells me that you really do need rcu_read_lock/unlock() for the guarantee here; the tasklist lock is not sufficient. The realtime kernel will preempt even those sections covered by spinlocks.
-- Dave
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