On 12:20, Dan Williams wrote:

+struct dma_async_tx_descriptor *
+async_r6_dd_recov(int disks, size_t bytes, int faila, int failb,
+		  struct page **ptrs, enum async_tx_flags flags,
+		  struct dma_async_tx_descriptor *depend_tx,
+		  dma_async_tx_callback cb, void *cb_param)
+{
+	struct dma_async_tx_descriptor *tx = NULL;
+	struct page *lptrs[disks];
+	unsigned char lcoef[disks-4];

This probably needs a BUG_ON(disks < 4).

+	 * B = (2^(y-x))*((2^(y-x) + {01})^(-1))

Minor optimization suggestion: As B depends only on y-x, there are
255 possible values for B, so a lookup table for all these values
would only occupy 255 bytes.

+ddr_sync:
+	{
+		void **sptrs = (void **)lptrs;

unnecessary cast

+struct dma_async_tx_descriptor *
+async_r6_dp_recov(int disks, size_t bytes, int faila, struct page **ptrs,
+		  enum async_tx_flags flags,
+		  struct dma_async_tx_descriptor *depend_tx,
+		  dma_async_tx_callback cb, void *cb_param)
+{
+	struct dma_async_tx_descriptor *tx = NULL;

unnecessary initialization.

+	struct page *lptrs[disks];
+	unsigned char lcoef[disks-2];
+	int i = 0, k = 0;

again. I'd suggest to init i and k in the for() loop.

Regards
Andre
-- 
The only person who always got his work done by Friday was Robinson Crusoe