qemu_img: is_not_zero() optimization
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>
diff --git a/qemu-img.c b/qemu-img.c
index 1da5484..4f162d1 100644
--- a/qemu-img.c
+++ b/qemu-img.c
@@ -496,14 +496,37 @@
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)
{
+ /*
+ * 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;
- len >>= 2;
- for(i = 0;i < len; i++) {
- if (((uint32_t *)sector)[i] != 0)
+ long d0, d1, d2, d3;
+ const long * const data = (const long *) sector;
+
+ 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 0;
}
