Messages in this thread | | | Subject | Re: BUG: MAX_STACK_TRACE_ENTRIES too low! | From | Bart Van Assche <> | Date | Mon, 11 Mar 2019 11:17:19 -0700 |
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On Mon, 2019-03-11 at 10:48 -0700, Linus Torvalds wrote: > On Mon, Mar 11, 2019 at 8:19 AM Bart Van Assche <bvanassche@acm.org> wrote: > > > > I think this issue has been fixed by a commit that went upstream yesterday. Hence: > > > > #syz fix: workqueue, lockdep: Fix an alloc_workqueue() error path > > Well, syzbot just reported a problem with that fix itself ("WARNING in > lockdep_unregister_key"). > > Looks like now the lockdep_unregister_key() cleanup might be called > even when the lockdep map was never initialized at all in that error > case. Hmm? > > I _think_ you need to split that "err_free_wq" label into "just free > the wq" and "unregister lockdep and free wq". > > But I didn't look any more closely, I might be misreading things.
Hi Linus,
From the console output of the syzbot bug report at the start of this e-mail thread:
WARNING: CPU: 1 PID: 7649 at kernel/locking/lockdep.c:747 register_lock_class+0x10de/0x2220
On line 747 in lockdep.c there is the following warning statement:
WARN_ON_ONCE(class->name != lock->name);
My interpretation is that this means that a lockdep key got reregistered without unregistration between the two registration calls. That's why I posted my "#syz fix: workqueue, lockdep: Fix an alloc_workqueue() error path" reply.
The latest syzbot complaint is different. In the console output of the latest report I found the following:
WARNING: CPU: 1 PID: 6970 at kernel/locking/lockdep.c:4925 lockdep_unregister_key+0x21c/0x4e0
On line 4925 there is the following warning statement:
WARN_ON_ONCE(!found);
In other words, lockdep_unregister_key() got called for a key that was never registered.
I agree with your conclusion that the err_free_wq label needs to be split. I will post a patch that realizes this.
Bart.
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