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

From: Christian Brauner
Date: Thu Dec 07 2023 - 16:29:16 EST


> 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.