[RFC PATCH v5 23/26] util: Add iova_tree_alloc

Peter Xu
Thu, 27 Jan 2022 11:25:22 -0000

On Thu, Jan 27, 2022 at 11:09:44AM +0100, Eugenio Perez Martin
wrote:
> > > +/**
> > > + * Try to accomodate a map of size ret->size in a hole
between
> > > + * max(end(hole_left), iova_start).
> >
> > I think this functions need the most comments, and above sentence is
more or
> > less not sounding correct... My try...
> >
> > /*
> >  * Try to find an unallocated IOVA range between LEFT and RIGHT
elements.
> >  *
> >  * There're three cases:
> >  *
> >  * (1) When LEFT==NULL, RIGHT must be non-NULL and it means we're
iterating at
> >  *     the 1st element.
> >  *
> >  * (2) When RIGHT==NULL, LEFT must be non-NULL and it means we're
iterating at
> >  *     the last element.
> >  *
> >  * (3) When both LEFT and RIGHT are non-NULL, this is the most common
case,
> >  *     we'll try to find a hole between LEFT and RIGHT mapping.
> >  */
> >
>
> This is also called with left == NULL and right == NULL in the first
> allocation with an empty tree. This allows iova_tree_alloc to have the
> same code path both if the three is empty or not.
>
> But I can add the use cases in the doc for sure.

Ah, right.

>
> > > + *
> > > + * @args Arguments to allocation
> > > + */
> > > +static bool iova_tree_alloc_map_in_hole(const struct
IOVATreeAllocArgs *args)
> > > +{
> > > +    const DMAMap *left = args->hole_left, *right =
args->hole_right;
> > > +    uint64_t hole_start, hole_last;
> > > +
> > > +    if (right && right->iova + right->size <
args->iova_begin) {
> > > +        return false;
> > > +    }
> > > +
> > > +    if (left && left->iova > args->iova_last) {
> > > +        return false;
> > > +    }
> > > +
> > > +    hole_start = MAX(left ? left->iova + left->size + 1 :
0, args->iova_begin);
> > > +    hole_last = MIN(right ? right->iova : HWADDR_MAX,
args->iova_last);
> >
> > I assume these values should be always inclusive, hence
> >
> > s/right->iova/right->iova + 1/
> >
> > ?
> >
>
> Right, it is confusing the way it's written. But I think it should be
> right->iova - 1 in any case to make it the hole last element, isn't
> it?

I was thinking "-1" but I failed to make it coherent with the thought
when
typing.. Heh.

>
> Would it work better to rename variable hole_last to hole_end? If not,
> we have a special case of the second allocation when iova_begin == 0:
> - We successfully allocate a DMAMap of size N. By the way the
> algorithm works,  it starts at 0, so [0, N] is allocated.

If we're always talking about inclusive ranges, shouldn't it be [0,
N-1]?

> - We try to allocate a second one of size M. At the first iteration,
> "right" is the previously allocated DMAMap.
> Using the -1 trick we get hole_end == HWADDR_MAX.

I'm not sure I get the point, but both naming look fine to me.  As long as
we
use inclusive ranges, then hole_end/last will be limited to HWADDR_MAX.

> > > +static gboolean iova_tree_alloc_traverse(gpointer key, gpointer
value,
> > > +                                         gpointer pargs)
> > > +{
> > > +    struct IOVATreeAllocArgs *args = pargs;
> > > +    DMAMap *node = value;
> > > +
> > > +    assert(key == value);
> > > +
> > > +    iova_tree_alloc_args_iterate(args, node);
> > > +    if (args->hole_left &&
args->hole_left->iova > args->iova_last) {
> >
> > IMHO this check is redundant and can be dropped, as it's already
done in
> > iova_tree_alloc_map_in_hole().
> >
>
> Assuming we add "iova_found" to iova_tree_alloc_map_in_hole to
> IOVATreeAllocArgs as you propose, it returns true if we are able to
> allocate a DMAMap entry, so no more iterations are needed. But if it
> returns false, it simply means that DMAMap cannot be allocated between
> left (or iova_begin) and right (iova_end). It doesn't tell if you can
> keep iterating or not. In other words, false == keep iterating if you
> can.
>
> This other check signals the end of the available hole, and to avoid
> iterating beyond iova_last in the (unlikely?) case we have more nodes
> to iterate beyond that.
>
> I'll try to make it more explicit.

Makes sense.  Comment works.

Thanks,

--
Peter Xu