Re: [RFC 1/3] pidfd: allow pidfd_open() on non-thread-group leaders

If these concerns are correct

So, ok. I misremebered this. The scenario I had been thinking of is
basically the following.

We have a thread-group with thread-group leader 1234 and a thread with
4567 in that thread-group. Assume current thread-group leader is tsk1
and the non-thread-group leader is tsk2. tsk1 uses struct pid *tg_pid
and tsk2 uses struct pid *t_pid. The struct pids look like this after
creation of both thread-group leader tsk1 and thread tsk2:

	TGID 1234				TID 4567 
	tg_pid[PIDTYPE_PID]  = tsk1		t_pid[PIDTYPE_PID]  = tsk2
	tg_pid[PIDTYPE_TGID] = tsk1		t_pid[PIDTYPE_TGID] = NULL

IOW, tsk2's struct pid has never been used as a thread-group leader and
thus PIDTYPE_TGID is NULL. Now assume someone does create pidfds for
tsk1 and for tsk2:
	
	tg_pidfd = pidfd_open(tsk1)		t_pidfd = pidfd_open(tsk2)
	-> tg_pidfd->private_data = tg_pid	-> t_pidfd->private_data = t_pid

So we stash away struct pid *tg_pid for a pidfd_open() on tsk1 and we
stash away struct pid *t_pid for a pidfd_open() on tsk2.

If we wait on that task via P_PIDFD we get:

				/* waiting through pidfd */
	waitid(P_PIDFD, tg_pidfd)		waitid(P_PIDFD, t_pidfd)
	tg_pid[PIDTYPE_TGID] == tsk1		t_pid[PIDTYPE_TGID] == NULL
	=> succeeds				=> fails

Because struct pid *tg_pid is used a thread-group leader struct pid we
can wait on that tsk1. But we can't via the non-thread-group leader
pidfd because the struct pid *t_pid has never been used as a
thread-group leader.

Now assume, t_pid exec's and the struct pids are transfered. IIRC, we
get:

	tg_pid[PIDTYPE_PID]   = tsk2		t_pid[PIDTYPE_PID]   = tsk1
	tg_pid[PIDTYPE_TGID]  = tsk2		t_pid[PIDTYPE_TGID]  = NULL

If we wait on that task via P_PIDFD we get:
	
				/* waiting through pidfd */
	waitid(P_PIDFD, tg_pidfd)		waitid(P_PIDFD, t_pid)
	tg_pid[PIDTYPE_TGID] == tsk2		t_pid[PIDTYPE_TGID] == NULL
	=> succeeds				=> fails

Which is what we want. So effectively this should all work and I
misremembered the struct pid linkage. So afaict we don't even have a
problem here which is great.