Patchwork [qemu-img] CPU consuming optimization

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Submitter Dmitry Konishchev
Date May 18, 2011, 10:27 a.m.
Message ID <4DD39EFB.8050400@gmail.com>
Download mbox | patch
Permalink /patch/96147/
State New
Headers show

Comments

Dmitry Konishchev - May 18, 2011, 10:27 a.m.
On 18.05.2011 13:31, Kevin Wolf wrote:
> Please move the declarations to the start of the function.
>
> I also would use a single line like "long d0, d1, d2, d3;", but that's
> up to you.
>
>> +
>> +    for(i = 0; i<  len; i += 4) {
>> +        d0 = ((const long*) sector)[i + 0];
>> +        d1 = ((const long*) sector)[i + 1];
>> +        d2 = ((const long*) sector)[i + 2];
>> +        d3 = ((const long*) sector)[i + 3];
>
> I would suggest to declare a const long* variable so that you don't have
> to cast each time you use, but that's probably a matter of taste.
>
>> +
>> +        if (d0 || d1 || d2 || d3)
>>                return 1;
>
> Coding style requires braces here.
>
>>        }
>> +
>>        return 0;
>>    }

OK, fixed.


Signed-off-by: Dmitry Konishchev <konishchev@gmail.com>
---
  qemu-img.c |   29 ++++++++++++++++++++++++++---
  1 files changed, 26 insertions(+), 3 deletions(-)

Patch

diff --git a/qemu-img.c b/qemu-img.c
index e825123..c849c6f 100644
--- a/qemu-img.c
+++ b/qemu-img.c
@@ -496,14 +496,37 @@  static int img_commit(int argc, char **argv)
      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;
  }