Thanks for cc'ing me...
On 04/26, Christian Brauner wrote:
On Thu, Apr 25, 2019 at 03:00:09PM -0400, Joel Fernandes (Google) wrote:
+static unsigned int pidfd_poll(struct file *file, struct poll_table_struct *pts) +{
- struct task_struct *task;
- struct pid *pid;
- int poll_flags = 0;
- /*
* tasklist_lock must be held because to avoid racing with* changes in exit_state and wake up. Basically to avoid:** P0: read exit_state = 0* P1: write exit_state = EXIT_DEAD* P1: Do a wake up - wq is empty, so do nothing* P0: Queue for polling - wait forever.*/- read_lock(&tasklist_lock);
- pid = file->private_data;
- task = pid_task(pid, PIDTYPE_PID);
- WARN_ON_ONCE(task && !thread_group_leader(task));
- if (!task || (task->exit_state && thread_group_empty(task)))
poll_flags = POLLIN | POLLRDNORM;
Joel, I still can't understand why do we need tasklist... and I don't really understand the comment. The code looks as if you are trying to avoid poll_wait(), but this would be strange.
OK, why can't pidfd_poll() do
poll_wait(file, &pid->wait_pidfd, pts);
rcu_read_lock(); task = pid_task(pid, PIDTYPE_PID); if (!task || task->exit_state && thread_group_empty(task)) poll_flags = POLLIN | ...; rcu_read_unlock();
return poll_flags;
?
+static void do_notify_pidfd(struct task_struct *task)
Maybe a short command that this helper can only be called when we know that task is a thread-group leader wouldn't hurt so there's no confusion later.
Not really. If the task is traced, do_notify_parent() (and thus do_notify_pidfd()) can be called to notify the debugger even if the task is not a leader and/or if it is not the last thread. The latter means a spurious wakeup for pidfd_poll().
+{
- struct pid *pid;
- lockdep_assert_held(&tasklist_lock);
- pid = get_task_pid(task, PIDTYPE_PID);
- wake_up_all(&pid->wait_pidfd);
- put_pid(pid);
Why get/put?
Oleg.