[AMD Official Use Only - General]

Hi Arun,
Thanks for your reply. comments are inline.
________________________________________
发件人: Paneer Selvam, Arunpravin <arunpravin.paneersel...@amd.com>
发送时间: 2022年11月29日 1:09
收件人: Pan, Xinhui; amd-gfx@lists.freedesktop.org
抄送: linux-ker...@vger.kernel.org; dri-de...@lists.freedesktop.org; 
matthew.a...@intel.com; dan...@ffwll.ch; Koenig, Christian
主题: Re: [PATCH v3] drm: Optimise for continuous memory allocation

Hi Xinhui,

On 11/28/2022 12:04 PM, xinhui pan wrote:
> Currently drm-buddy does not have full knowledge of continuous memory.
>
> Lets consider scenario below.
> order 1:    L             R
> order 0: LL   LR      RL      RR
> for order 1 allocation, it can offer L or R or LR+RL.
>
> For now, we only implement L or R case for continuous memory allocation.
> So this patch aims to implement the LR+RL case.
>
> Signed-off-by: xinhui pan <xinhui....@amd.com>
> ---
> change from v2:
> search continuous block in nearby root if needed
>
> change from v1:
> implement top-down continuous allocation
> ---
>   drivers/gpu/drm/drm_buddy.c | 78 +++++++++++++++++++++++++++++++++----
>   1 file changed, 71 insertions(+), 7 deletions(-)
>
> diff --git a/drivers/gpu/drm/drm_buddy.c b/drivers/gpu/drm/drm_buddy.c
> index 11bb59399471..ff58eb3136d2 100644
> --- a/drivers/gpu/drm/drm_buddy.c
> +++ b/drivers/gpu/drm/drm_buddy.c
> @@ -386,6 +386,58 @@ alloc_range_bias(struct drm_buddy *mm,
>       return ERR_PTR(err);
>   }
>
> +static struct drm_buddy_block *
> +find_continuous_blocks(struct drm_buddy *mm,
> +                    int order,
> +                    unsigned long flags,
> +                    struct drm_buddy_block **rn)
> +{
> +     struct list_head *head = &mm->free_list[order];
> +     struct drm_buddy_block *node, *parent, *free_node, *max_node = NULL;
NIT: We usually name the variable as *block or ***_block for drm buddy
and we have *node or ***_node for drm mm manager.

[xh] Oh, yes. The code naming is important. Will fix it.

> +     int i;
> +
> +     list_for_each_entry(free_node, head, link) {
> +             if (max_node) {
> +                     if (!(flags & DRM_BUDDY_TOPDOWN_ALLOCATION))
> +                             break;
> +
> +                     if (drm_buddy_block_offset(free_node) <
> +                         drm_buddy_block_offset(max_node))
> +                             continue;
> +             }
> +
> +             parent = free_node;
> +             do {
> +                     node = parent;
> +                     parent = parent->parent;
> +             } while (parent && parent->right == node);
> +
> +             if (!parent) {
> +                     for (i = 0; i < mm->n_roots - 1; i++)
> +                             if (mm->roots[i] == node)
> +                                     break;
> +                     if (i == mm->n_roots - 1)
> +                             continue;
> +                     node = mm->roots[i + 1];
> +             } else {
> +                     node = parent->right;
> +             }
> +
> +             while (drm_buddy_block_is_split(node))
> +                     node = node->left;
> +
> +             if (drm_buddy_block_is_free(node) &&
> +                 drm_buddy_block_order(node) == order) {
> +                     *rn = node;
> +                     max_node = free_node;
> +                     BUG_ON(drm_buddy_block_offset(node) !=
> +                             drm_buddy_block_offset(max_node) +
> +                             drm_buddy_block_size(mm, max_node));
> +             }
> +     }
> +     return max_node;
> +}
> +
>   static struct drm_buddy_block *
>   get_maxblock(struct list_head *head)
>   {
> @@ -637,7 +689,7 @@ int drm_buddy_alloc_blocks(struct drm_buddy *mm,
>                          struct list_head *blocks,
>                          unsigned long flags)
>   {
> -     struct drm_buddy_block *block = NULL;
> +     struct drm_buddy_block *block = NULL, *rblock = NULL;
>       unsigned int min_order, order;
>       unsigned long pages;
>       LIST_HEAD(allocated);
> @@ -689,17 +741,29 @@ int drm_buddy_alloc_blocks(struct drm_buddy *mm,
>                               break;
>
>                       if (order-- == min_order) {
> +                             if (!(flags & DRM_BUDDY_RANGE_ALLOCATION) &&
> +                                 min_order != 0 && pages == BIT(order + 1)) {
> +                                     block = find_continuous_blocks(mm,
> +                                                                    order,
> +                                                                    flags,
> +                                                                    &rblock);
> +                                     if (block)
> +                                             break;
> +                             }
>                               err = -ENOSPC;
>                               goto err_free;
>                       }
>               } while (1);
>
> -             mark_allocated(block);
> -             mm->avail -= drm_buddy_block_size(mm, block);
> -             kmemleak_update_trace(block);
> -             list_add_tail(&block->link, &allocated);
> -
> -             pages -= BIT(order);
> +             do {
> +                     mark_allocated(block);
> +                     mm->avail -= drm_buddy_block_size(mm, block);
> +                     kmemleak_update_trace(block);
> +                     list_add_tail(&block->link, &allocated);
> +                     pages -= BIT(order);
> +                     block = rblock;
> +                     rblock = NULL;
> +             } while (block);
I think with this approach, if we are lucky enough we may get contiguous
blocks in one order level down in RL
combination from the freelist?

[xh] That is just what I did at first time like something below.
       list_for_each_entry(node, head, link)
               list_for_each_entry_reverse(rnode, head, link)

I do the test with one ROCM application while running gdm restart background 
which is a very normal scenario.

The test result shows there is about 4% chance to find the continuous blocks in 
this case. This 4% is really good enough as eviction is much more expensive.
So after that, I need take care of the rest 96% to not introduce too much 
workload.

Compared with the two loops, walking through the tree is a little cheaper.

thanks
xinhui

Regards,
Arun
>
>               if (!pages)
>                       break;

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