i40e/i40evf: avoid atomics

In the case where we have a page fully used by receive data, we need to
release the page fully to the stack. Instead of calling get_page (which
increments the page count) followed by free_page (which decrements the
page count), just donate our reference to the stack. Although this
donation is not tax deductible, it does allow us to avoid two very
expensive atomic operations that reverse each other.

Change-ID: If70739792d5748995fc175ec92ac2171ed4ad8fc
Signed-off-by: NMitch Williams <mitch.a.williams@intel.com>
Tested-by: NAndrew Bowers <andrewx.bowers@intel.com>
Signed-off-by: NJeff Kirsher <jeffrey.t.kirsher@intel.com>

drivers/net/ethernet/intel/i40e/i40e_txrx.c

@@ -1663,28 +1663,33 @@ static int i40e_clean_rx_irq_ps(struct i40e_ring *rx_ring, const int budget)
 					rx_bi->page_offset + copysize,
 					rx_packet_len, I40E_RXBUFFER_2048);
 
-			get_page(rx_bi->page);
-			/* switch to the other half-page here; the allocation
-			 * code programs the right addr into HW. If we haven't
-			 * used this half-page, the address won't be changed,
-			 * and HW can just use it next time through.
-			 */
-			rx_bi->page_offset ^= PAGE_SIZE / 2;
 			/* If the page count is more than 2, then both halves
 			 * of the page are used and we need to free it. Do it
 			 * here instead of in the alloc code. Otherwise one
 			 * of the half-pages might be released between now and
 			 * then, and we wouldn't know which one to use.
+			 * Don't call get_page and free_page since those are
+			 * both expensive atomic operations that just change
+			 * the refcount in opposite directions. Just give the
+			 * page to the stack; he can have our refcount.
 			 */
 			if (page_count(rx_bi->page) > 2) {
 				dma_unmap_page(rx_ring->dev,
 					       rx_bi->page_dma,
 					       PAGE_SIZE,
 					       DMA_FROM_DEVICE);
-				__free_page(rx_bi->page);
 				rx_bi->page = NULL;
 				rx_bi->page_dma = 0;
 				rx_ring->rx_stats.realloc_count++;
+			} else {
+				get_page(rx_bi->page);
+				/* switch to the other half-page here; the
+				 * allocation code programs the right addr
+				 * into HW. If we haven't used this half-page,
+				 * the address won't be changed, and HW can
+				 * just use it next time through.
+				 */
+				rx_bi->page_offset ^= PAGE_SIZE / 2;
 			}
 
 		}

drivers/net/ethernet/intel/i40evf/i40e_txrx.c

@@ -1126,28 +1126,33 @@ static int i40e_clean_rx_irq_ps(struct i40e_ring *rx_ring, const int budget)
 					rx_bi->page_offset + copysize,
 					rx_packet_len, I40E_RXBUFFER_2048);
 
-			get_page(rx_bi->page);
-			/* switch to the other half-page here; the allocation
-			 * code programs the right addr into HW. If we haven't
-			 * used this half-page, the address won't be changed,
-			 * and HW can just use it next time through.
-			 */
-			rx_bi->page_offset ^= PAGE_SIZE / 2;
 			/* If the page count is more than 2, then both halves
 			 * of the page are used and we need to free it. Do it
 			 * here instead of in the alloc code. Otherwise one
 			 * of the half-pages might be released between now and
 			 * then, and we wouldn't know which one to use.
+			 * Don't call get_page and free_page since those are
+			 * both expensive atomic operations that just change
+			 * the refcount in opposite directions. Just give the
+			 * page to the stack; he can have our refcount.
 			 */
 			if (page_count(rx_bi->page) > 2) {
 				dma_unmap_page(rx_ring->dev,
 					       rx_bi->page_dma,
 					       PAGE_SIZE,
 					       DMA_FROM_DEVICE);
-				__free_page(rx_bi->page);
 				rx_bi->page = NULL;
 				rx_bi->page_dma = 0;
 				rx_ring->rx_stats.realloc_count++;
+			} else {
+				get_page(rx_bi->page);
+				/* switch to the other half-page here; the
+				 * allocation code programs the right addr
+				 * into HW. If we haven't used this half-page,
+				 * the address won't be changed, and HW can
+				 * just use it next time through.
+				 */
+				rx_bi->page_offset ^= PAGE_SIZE / 2;
 			}
 
 		}