Re: [RFC] resource: Avoid unnecessary resource tree walking in __region_intersects()

[David Hildenbrand]

On 10.10.24 08:55, Huang Ying wrote:
> Currently, if __region_intersects() finds any overlapped but unmatched
> resource, it walks the descendant resource tree to check for
> overlapped and matched descendant resources.  This is achieved using
> for_each_resource(), which iterates not only the descent tree, but
> also subsequent sibling trees in certain scenarios.  While this
> doesn't introduce bugs, it makes code hard to be understood and
> potentially inefficient.
>
> So, the patch renames next_resource() to __next_resource() and
> modified it to return NULL after traversing all descent resources.
> Test shows that this avoids unnecessary resource tree walking in
> __region_intersects().
>
> It appears even better to revise for_each_resource() to traverse the
> descendant resource tree of "_root" only.  But that will cause "_root"
> to be evaluated twice, which I don't find a good way to eliminate.

I'm not sure I'm enjoying below code, it makes it harder for me to
understand what's happening.

I'm also not 100% sure why "p" becomes "root" and "dp" becomes "p" when
calling the function :) Likely this works as intended, but it's confusing
(IOW, bad naming, especially for dp).


I think you should just leave next_resource() alone and rather add
a new function that doesn't conditionally consume NULL pointers
(and also no skip_children because you're passing false either way).

static struct resource *next_resource_XXX(struct resource *root,
		struct resource *p)
{
	while (!p->sibling && p->parent) {
		p = p->parent;
		if (p == root)
			return NULL;
	}
	return p->sibling;
}

Maybe even better, add a new for_each_resource() macro that expresses the intended semantics.

#define for_each_resource_XXX(_root, _p) \
	for ((_p) = (_root)->child; (_p); (_p) = next_resource_XXX(_root, _p))

XXX TBD

Or do you think this should not only be "improved" for the __region_intersects() use case
but for all for_each_resource() users? I cannot tell easily.

--
Cheers,

David / dhildenb