From mboxrd@z Thu Jan 1 00:00:00 1970 From: Theodore Dubois Subject: edge case posix file lock deadlock detection Date: Wed, 30 Jan 2019 19:31:10 -0800 Message-ID: <8E383B31-AB1C-449E-8AC4-DE20228CA8FF@icloud.com> Mime-Version: 1.0 (Mac OS X Mail 11.5 \(3445.9.1\)) Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Return-path: List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Errors-To: kernelnewbies-bounces@kernelnewbies.org To: Theodore Dubois Cc: linux-man@vger.kernel.org, kernelnewbies@kernelnewbies.org List-Id: linux-man@vger.kernel.org I'm having trouble figuring out how the kernel handles a particular case in deadlock detection on posix file locks. Here's the scenario: PID 1: locks byte 2 PID 3: locks byte 0 PID 2: locks byte 10 PID 1: locks byte 10 PID 2: locks bytes 0-2 inclusive The last step fails with EDEADLK, but I'm not sure how that is detected. The specified range conflicts with two different locks, and the first loop in posix_lock_inode would find whichever one comes first in the linked list, and pass that to the deadlock detector. If the lock on byte 2 comes first in the list, a cycle would be found between the lock on byte 2 and the lock on byte 10. But if the lock on byte 0 comes first, the deadlock detector would return NULL from what_owner_is_waiting_for on that lock since PID 3 has no other locks, and immediately succeed. What am I missing? Thanks, ~Theodore