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SubjectRe: [RESEND PATCH 01/11] kernel:lockdep:print the shortest dependency chain if finding a circle

* Ming Lei <tom.leiming@gmail.com> wrote:

> > If I understand well, lets imagine the following:
> >
> > Task 1 acquires: A B F
> > Task 2 acquires: A B C D E F
> > Task 3 acquires: F B
> >
> > Before your patch, DFS would report the BF - FB wicked
> > dependency by reporting the Task 2 dependency snapshot, which is
> > cluttered by the other locks C, D and E (that would be reported
> > in the dependency chain if I'm not wrong) whereas BFS would be
> > smarter by finding the shortest snapshot found in Task 3: just F
> > B.
> >
> > Correct me if I misunderstand this patch.
>
> You are correct, BFS will always find the shortest circle if there
> is circle.
>
> > I suggest you to provide an example along this patch, that would
> > make it easier to demonstrate its importance.
>
> No, as you said, the shortest circle is not very important, and it
> is just a byproduct and we have no reason to reject it.

It's a nice byproduct, beyond the primary advantage of not being a
stack based recursion check.

I think this patch-set is great, and there's just one more step
needed to make it round: it would be nice to remove the limitation
of maximum number of locks held per task. (MAX_LOCK_DEPTH)

The way we could do it is to split out this bit of struct task:

#ifdef CONFIG_LOCKDEP
# define MAX_LOCK_DEPTH 48UL
u64 curr_chain_key;
int lockdep_depth;
unsigned int lockdep_recursion;
struct held_lock held_locks[MAX_LOCK_DEPTH];
gfp_t lockdep_reclaim_gfp;
#endif

into a separate 'struct lockdep_state' structure, and allocate it
dynamically during fork with a initial pre-set size of say 64 locks
depth. If we hit that limit, we'd double the allocation threshold,
which would cause a larger structure to be allocated for all newly
allocated tasks.

( This means that the task that hits this threshold needs to have
lockdep disabled until it exits - but that's OK. )

Ingo


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