ptp: remove ptp->n_vclocks check logic in ptp_vclock_in_use()
In the Linux kernel, the following vulnerability has been resolved:
ptp: remove ptp->n_vclocks check logic in ptp_vclock_in_use()
There is no disagreement that we should check both ptp->is_virtual_clock
and ptp->n_vclocks to check if the ptp virtual clock is in use.
However, when we acquire ptp->n_vclocks_mux to read ptp->n_vclocks in
ptp_vclock_in_use(), we observe a recursive lock in the call trace
starting from n_vclocks_store().
============================================
WARNING: possible recursive locking detected
6.15.0-rc6 #1 Not tainted
--------------------------------------------
syz.0.1540/13807 is trying to acquire lock:
ffff888035a24868 (&ptp->n_vclocks_mux){+.+.}-{4:4}, at:
ptp_vclock_in_use drivers/ptp/ptp_private.h:103 [inline]
ffff888035a24868 (&ptp->n_vclocks_mux){+.+.}-{4:4}, at:
ptp_clock_unregister+0x21/0x250 drivers/ptp/ptp_clock.c:415
but task is already holding lock:
ffff888030704868 (&ptp->n_vclocks_mux){+.+.}-{4:4}, at:
n_vclocks_store+0xf1/0x6d0 drivers/ptp/ptp_sysfs.c:215
other info that might help us debug this:
Possible unsafe locking scenario:
CPU0
----
lock(&ptp->n_vclocks_mux);
lock(&ptp->n_vclocks_mux);
* DEADLOCK *
....
============================================
The best way to solve this is to remove the logic that checks
ptp->n_vclocks in ptp_vclock_in_use().
The reason why this is appropriate is that any path that uses
ptp->n_vclocks must unconditionally check if ptp->n_vclocks is greater
than 0 before unregistering vclocks, and all functions are already
written this way. And in the function that uses ptp->n_vclocks, we
already get ptp->n_vclocks_mux before unregistering vclocks.
Therefore, we need to remove the redundant check for ptp->n_vclocks in
ptp_vclock_in_use() to prevent recursive locking.
