ptp: remove ptp->n_vclocks check logic in ptp_vclock_in_use()

Summary

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.