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qemu_img: is_not_zero() optimization

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I run qemu-img under profiler and realized, that most of CPU time is
consumed by is_not_zero() function. I had made a couple of optimizations
on it and got the following output for `time qemu-img convert -O qcow2
volume.qcow2 snapshot.qcow2`:

Original qemu-img:
real 0m56.159s
user 0m34.670s
sys  0m12.079s

Patched qemu-img:
real 0m34.805s
user 0m18.445s
sys  0m12.552s

Signed-off-by: Dmitry Konishchev <konishchev@gmail.com>
Signed-off-by: Kevin Wolf <kwolf@redhat.com>
This commit is contained in:
Dmitry Konishchev 2011-05-18 15:03:59 +04:00 committed by Kevin Wolf
parent 6f321e93ab
commit f6a00aa150
1 changed files with 26 additions and 3 deletions
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29
qemu-img.c
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@ -496,14 +496,37 @@ static int img_commit(int argc, char **argv)
return 0; return 0;
} }
/*
* Checks whether the sector is not a zero sector.
*
* Attention! The len must be a multiple of 4 * sizeof(long) due to
* restriction of optimizations in this function.
*/
static int is_not_zero(const uint8_t *sector, int len) static int is_not_zero(const uint8_t *sector, int len)
{ {
/*
* Use long as the biggest available internal data type that fits into the
* CPU register and unroll the loop to smooth out the effect of memory
* latency.
*/
int i; int i;
len >>= 2; long d0, d1, d2, d3;
for(i = 0;i < len; i++) { const long * const data = (const long *) sector;
if (((uint32_t *)sector)[i] != 0)
len /= sizeof(long);
for(i = 0; i < len; i += 4) {
d0 = data[i + 0];
d1 = data[i + 1];
d2 = data[i + 2];
d3 = data[i + 3];
if (d0 || d1 || d2 || d3) {
return 1; return 1;
}
} }
return 0; return 0;
} }