vulnerability
Debian: CVE-2025-39977: linux, linux-6.1 -- security update
| Severity | CVSS | Published | Added | Modified |
|---|---|---|---|---|
| 4 | (AV:L/AC:M/Au:S/C:N/I:N/A:C) | Nov 13, 2025 | Nov 13, 2025 | Dec 17, 2025 |
Description
In the Linux kernel, the following vulnerability has been resolved: futex: Prevent use-after-free during requeue-PI syzbot managed to trigger the following race: T1 T2 futex_wait_requeue_pi() futex_do_wait() schedule() futex_requeue() futex_proxy_trylock_atomic() futex_requeue_pi_prepare() requeue_pi_wake_futex() futex_requeue_pi_complete() /* preempt */ * timeout/ signal wakes T1 * futex_requeue_pi_wakeup_sync() // Q_REQUEUE_PI_LOCKED futex_hash_put() // back to userland, on stack futex_q is garbage /* back */ wake_up_state(q->task, TASK_NORMAL); In this scenario futex_wait_requeue_pi() is able to leave without using futex_q::lock_ptr for synchronization. This can be prevented by reading futex_q::task before updating the futex_q::requeue_state. A reference on the task_struct is not needed because requeue_pi_wake_futex() is invoked with a spinlock_t held which implies a RCU read section. Even if T1 terminates immediately after, the task_struct will remain valid during T2's wake_up_state(). A READ_ONCE on futex_q::task before futex_requeue_pi_complete() is enough because it ensures that the variable is read before the state is updated. Read futex_q::task before updating the requeue state, use it for the following wakeup.
Solutions
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