Thread (12 messages) read the whole thread 12 messages, 4 authors, 2018-04-13

[PATCH v2 1/2] dmaengine: stm32-mdma: align TLEN and buffer length on burst

From: robin.murphy@arm.com (Robin Murphy)
Date: 2018-04-13 11:09:40
Also in: dmaengine, lkml

On 13/04/18 10:45, Pierre Yves MORDRET wrote:
Hi Robin

On 04/11/2018 05:14 PM, Robin Murphy wrote:
quoted
On 11/04/18 15:44, Pierre-Yves MORDRET wrote:
quoted
Both buffer Transfer Length (TLEN if any) and transfer size have to be
aligned on burst size (burst beats*bus width).

Signed-off-by: Pierre-Yves MORDRET <redacted>
---
    Version history:
      v1:
         * Initial
      v2:
---
---
   drivers/dma/stm32-mdma.c | 2 +-
   1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/drivers/dma/stm32-mdma.c b/drivers/dma/stm32-mdma.c
index daa1602..fbcffa2 100644
--- a/drivers/dma/stm32-mdma.c
+++ b/drivers/dma/stm32-mdma.c
@@ -413,7 +413,7 @@ static u32 stm32_mdma_get_best_burst(u32 buf_len, u32 tlen, u32 max_burst,
   	u32 best_burst = max_burst;
   	u32 burst_len = best_burst * width;
   
-	while ((burst_len > 0) && (tlen % burst_len)) {
+	while ((burst_len > 0) && (((tlen | buf_len) & (burst_len - 1)) != 0)) {
   		best_burst = best_burst >> 1;
   		burst_len = best_burst * width;
   	}
FWIW, doesn't that whole loop come down to just:

	burst_len = min(ffs(tlen | buf_len), max_burst * width);
No sure it ends as expected. or I miss something or don't understand this statement
I tried with "relevant value" : i.e. best_burst = 32, Tlen=128(default) and
buf_len = 64, width= 4. This statements gets me something wrong output => 7
instead of 16 * 4.
I doubt :)
Heh, seems I confused myself halfway through and started thinking 
max_burst and width were the exponents x rather than the values 2^x...

A more representative guess should be:

	min(1 << __ffs(tlen | buf_len), max_burst * width);

but the general point I was trying to make is that a loop checking 
whether the bottom n bits of something are zero for different values of 
n is unnecessary when n can simply be calculated directly*.

Robin.


* in the case of this "just the lowest set bit" idiom there's also the 
shift-free ((x & (x - 1)) ^ x), but as well as being unreadable it's 
generally less efficient than (1 << __ffs(x)) for most modern ISAs.
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