On 9/30/2020 2:58 PM, Jason Gunthorpe wrote:
On Wed, Sep 30, 2020 at 02:53:58PM +0300, Maor Gottlieb wrote:
On 9/30/2020 2:45 PM, Jason Gunthorpe wrote:
On Wed, Sep 30, 2020 at 12:53:21PM +0300, Leon Romanovsky wrote:
On Tue, Sep 29, 2020 at 04:59:29PM -0300, Jason Gunthorpe wrote:
On Sun, Sep 27, 2020 at 09:46:47AM +0300, Leon Romanovsky wrote:
@@ -296,11 +223,17 @@ static struct ib_umem *__ib_umem_get(struct ib_device *device, goto umem_release;
cur_base += ret * PAGE_SIZE;
npages -= ret;sg = ib_umem_add_sg_table(sg, page_list, ret,dma_get_max_seg_size(device->dma_device),&umem->sg_nents);
npages -= ret;sg = __sg_alloc_table_from_pages(&umem->sg_head, page_list, ret, 0, ret << PAGE_SHIFT,dma_get_max_seg_size(device->dma_device), sg, npages,GFP_KERNEL);umem->sg_nents = umem->sg_head.nents;if (IS_ERR(sg)) {unpin_user_pages_dirty_lock(page_list, ret, 0);ret = PTR_ERR(sg);goto umem_release;}}
sg_mark_end(sg);
Does it still need the sg_mark_end?
It is preserved here for correctness, the release logic doesn't rely on this marker, but it is better to leave it.
I mean, my read of __sg_alloc_table_from_pages() is that it already placed it, the final __alloc_table() does it?
It marks the last allocated sge, but not the last populated sge (with page).
Why are those different?
It looks like the last iteration calls __alloc_table() with an exact number of sges
- if (!prv) {
/* Only the last allocation could be less than the maximum */table_size = left_pages ? SG_MAX_SINGLE_ALLOC : chunks;ret = sg_alloc_table(sgt, table_size, gfp_mask);if (unlikely(ret))return ERR_PTR(ret);- }
Jason
This is right only for the last iteration. E.g. in the first iteration in case that there are more pages (left_pages), then we allocate SG_MAX_SINGLE_ALLOC. We don't know how many pages from the second iteration will be squashed to the SGE from the first iteration.