Messages in this thread | | | Subject | Re: [PATCH v2 1/2] dmaengine: stm32-mdma: align TLEN and buffer length on burst | From | Robin Murphy <> | Date | Fri, 13 Apr 2018 12:09:35 +0100 |
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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@st.com> >>> --- >>> 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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