Entreprise française de cybersécurité depuis 2004
☎ 03 60 47 09 81 - info@securiteinfo.com
☎ 03 60 47 09 81 - info@securiteinfo.com
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_clockand 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 inptp_vclock_in_use(), we observe a recursive lock in the call tracestarting from n_vclocks_store().============================================WARNING: possible recursive locking detected6.15.0-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:415but task is already holding lock:ffff888030704868 (&ptp->n_vclocks_mux){+.+.}-{4:4}, at: n_vclocks_store+0xf1/0x6d0 drivers/ptp/ptp_sysfs.c:215other 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 checksptp->n_vclocks in ptp_vclock_in_use().The reason why this is appropriate is that any path that usesptp->n_vclocks must unconditionally check if ptp->n_vclocks is greaterthan 0 before unregistering vclocks, and all functions are alreadywritten this way. And in the function that uses ptp->n_vclocks, wealready get ptp->n_vclocks_mux before unregistering vclocks.Therefore, we need to remove the redundant check for ptp->n_vclocks inptp_vclock_in_use() to prevent recursive locking.
No PoCs from references.
- https://github.com/w4zu/Debian_security