On 04/13/2018 01:09 PM, Robin Murphy wrote:
> On 13/04/18 10:45, Pierre Yves MORDRET wrote:
>> Hi Robin
>>
>> On 04/11/2018 05:14 PM, Robin Murphy wrote:
>>> On 11/04/18 15:44, Pierre-Yves MORDRET wrote:
>>>> 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 <pierre-yves.mordret@xxxxxx>
>>>> ---
>>>> 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.
Got the point. I figure how to compute this value with __ffs. Your last
statement doesn't provide the good value, but the spirit is here. I just have to
adjust with what I want.
Thx
>
>
> * 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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