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.