> Am 05.07.2018 um 17:15 schrieb Kevin Wolf <kw...@redhat.com>: > > Am 05.07.2018 um 12:52 hat Peter Lieven geschrieben: >> We currently don't enforce that the sparse segments we detect during convert >> are >> aligned. This leads to unnecessary and costly read-modify-write cycles either >> internally in Qemu or in the background on the storage device as nearly all >> modern filesystems or hardware have a 4k alignment internally. >> >> The number of RMW cycles when converting an example image [1] to a raw >> device that >> has 4k sector size is about 4600 4k read requests to perform a total of >> about 15000 >> write requests. With this path the additional 4600 read requests are >> eliminated. >> >> [1] >> https://cloud-images.ubuntu.com/releases/16.04/release/ubuntu-16.04-server-cloudimg-amd64-disk1.vmdk >> >> Signed-off-by: Peter Lieven <p...@kamp.de> >> --- >> V2->V3: - ensure that s.alignment is a power of 2 >> - correctly handle n < alignment in is_allocated_sectors if >> sector_num % alignment > 0. >> V1->V2: - take the current sector offset into account [Max] >> - try to figure out the target alignment [Max] >> >> qemu-img.c | 46 ++++++++++++++++++++++++++++++++++++---------- >> 1 file changed, 36 insertions(+), 10 deletions(-) >> >> diff --git a/qemu-img.c b/qemu-img.c >> index e1a506f..db91b9e 100644 >> --- a/qemu-img.c >> +++ b/qemu-img.c >> @@ -1105,8 +1105,11 @@ static int64_t find_nonzero(const uint8_t *buf, >> int64_t n) >> * >> * 'pnum' is set to the number of sectors (including and immediately >> following >> * the first one) that are known to be in the same allocated/unallocated >> state. >> + * The function will try to align 'pnum' to the number of sectors specified >> + * in 'alignment' to avoid unnecassary RMW cycles on modern hardware. >> */ >> -static int is_allocated_sectors(const uint8_t *buf, int n, int *pnum) >> +static int is_allocated_sectors(const uint8_t *buf, int n, int *pnum, >> + int64_t sector_num, int alignment) >> { >> bool is_zero; >> int i; >> @@ -1115,14 +1118,26 @@ static int is_allocated_sectors(const uint8_t *buf, >> int n, int *pnum) >> *pnum = 0; >> return 0; >> } >> - is_zero = buffer_is_zero(buf, 512); >> - for(i = 1; i < n; i++) { >> - buf += 512; >> - if (is_zero != buffer_is_zero(buf, 512)) { >> + >> + if (n % alignment) { >> + alignment = 1; >> + } > > So if n is unaligned, we keep the result unaligned, too. Makes sense, > because otherwise we'd just split the request in two, but still get the > same result. > > Worth mentioning in the function comment, though? > >> + >> + if (sector_num % alignment) { >> + n = ROUND_UP(sector_num, alignment) - sector_num; >> + alignment = 1; >> + } > > So if the start is unaligned, only check until the next alignment > boundary. > > This one isn't obvious to me. Doesn't it result in the same scenario > where a request is needlessly split in two? Wouldn't it be better to > first check the unaligned head and then continue with the rest of n if > that results in an aligned end offset? > > Actually, should the order of both checks be reversed, because an > unaligned n with an unaligned sector_num could actually result in an > aligned end offset?
I think the key is that the function should just try to end with an aligned offset. It seems not trivial to decide how to round up or down to the next boundary tough... I think for the allocated case its just rounding up to the next boundary (if its not beyond the end of the buffer). If the status is unallocated it is rounding down, if *pnum will still be > 0? Please keep in mind that is_allocated_sectors is only invoked from is_allocated_sectors_min, so a small *pnum for unallocated sectors will have no effect of a split request as everything below min (which is min_sparse) will be count as allocated. Best, Peter
