Re: [PATCH 2/6] locking: Introduce range reader/writer lock

[Davidlohr Bueso <dave@stgolabs.net>]

On Mon, 15 May 2017, Peter Zijlstra wrote:

>On Mon, May 15, 2017 at 02:07:21AM -0700, Davidlohr Bueso wrote:
>
>> + * Fairness and freedom of starvation are guaranteed by the lack of lock
>> + * stealing, thus range locks depend directly on interval tree semantics.
>> + * This is particularly for iterations, where the key for the rbtree is
>> + * given by the interval's low endpoint,
>
>
>So suppose the lock is held at [a,n], and I want to acquire [g,z], this
>conflicts, therefore I wait.

Ok, then, ref [g,z] = 1, ref [a,n] = 0 (lock owner). Per below,
at this point the tree will overlap with anything between [a,z],
which is the world.

>
>While I wait, someone else comes in at [b,m], they too wait.

[b,m] intersects with both nodes above, thus ref [b,m] = 2.

>
>[a,n] is released, per ordering [b,m] acquires, I still wait.

Now:
ref [g,z] = 0
ref [b,m] = 1

So due to reference counting [g,z] is acquired, despite [b,m]
being _put() before [g,z].

>[a,n] returns to wait.

Similar to the [b,m] case, when [a,n] comes back, it too will
get a ref = 2 and hence "go back in line". Iow, lock order does
depend on have fifo semantics among contended ranges.

>
>[b,m] releases, does the iteration then restart and grant it to [a,n] or
>will I (at [g,z]) finally acquire?
>
>
>Since the code always does range_interval_tree_foreach() it would appear
>to me [b,m] will always win and [g,z] could be made to wait
>indefinitely (by always contending with another range that has a lower
>starting point).
>
>
>
>>                                          and duplicates are walked as it
>> + * would an inorder traversal of the tree.
>
>Are duplicates ordered in FIFO ? Afaict the above is free of actual
>semantics.

This will strictly depend on the rotation when you have duplicates
when more nodes are added later. But again that's the order of
walking the tree.

Thanks,
Davidlohr