Re: [patch 1/3] radix tree: RCU lockless read-side

[Nick Piggin <nickpiggin@yahoo.com.au>]

Balbir Singh wrote:

> <snip>
>
> > But we should have already rcu_dereference()ed "slot", right
> > (in the loop above this one)? That means we are now able to
> > dereference it, and the data at the other end will be valid.
>
> Yes, but my confusion is about the following piece of code
>
> <begin code>
>
>       for ( ; height > 1; height--) {
>
>               for (i = (index >> shift) & RADIX_TREE_MAP_MASK ;
>                               i < RADIX_TREE_MAP_SIZE; i++) {
> -                       if (slot->slots[i] != NULL)
> +                       __s = rcu_dereference(slot->slots[i]);
> +                       if (__s != NULL)
>                               break;
>                       index &= ~((1UL << shift) - 1);
>                       index += 1UL << shift;
> @@ -531,14 +550,14 @@ __lookup(struct radix_tree_root *root, v
>                       goto out;
>
>               shift -= RADIX_TREE_MAP_SHIFT;
> -               slot = slot->slots[i];
> +               slot = __s;
>       }
>
>       /* Bottom level: grab some items */
>       for (i = index & RADIX_TREE_MAP_MASK; i < RADIX_TREE_MAP_SIZE; i++) {
>               index++;
>               if (slot->slots[i]) {
> -                       results[nr_found++] = slot->slots[i];
> +                       results[nr_found++] = &slot->slots[i];
>                       if (nr_found == max_items)
>                               goto out;
>               }
> <end code>
>
> In the for loop, lets say __s is *not* NULL, we break from the loop.
> In the loop below
> slot->slots[i] is derefenced without rcu, __s is not used. Is that not
> inconsistent?

The "slots" member is an array, not an RCU assigned pointer. As such, after
doing rcu_dereference(slot), you can access slot->slots[i] without further
memory barriers I think?

But I agree that code now is a bit inconsistent. I've cleaned things up a
bit in my tree now... but perhaps it is easier if you send a patch to show
what you mean (because sometimes I'm a bit dense, I'm afraid).

Thanks,
Nick

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