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authorJoel Becker <joel.becker@oracle.com>2008-12-15 18:24:33 -0800
committerMark Fasheh <mfasheh@suse.com>2009-01-05 08:40:35 -0800
commit7bb458a58588f397068e4166c615e9fcc7480c16 (patch)
treea5d330c12f983837097f775e2d7c6c4d85fb9acd /fs
parente798b3f8a920c82a8e556dd54df97f0d3d0f9144 (diff)
ocfs2: Another hamming code optimization.
In the calc_code_bit() function, we must find all powers of two beneath the code bit number, *after* it's shifted by those powers of two. This requires a loop to see where it ends up. We can optimize it by starting at its most significant bit. This shaves 32% off the time, for a total of 67.6% shaved off of the original, naive implementation. Signed-off-by: Joel Becker <joel.becker@oracle.com> Signed-off-by: Mark Fasheh <mfasheh@suse.com>
Diffstat (limited to 'fs')
-rw-r--r--fs/ocfs2/blockcheck.c40
1 files changed, 39 insertions, 1 deletions
diff --git a/fs/ocfs2/blockcheck.c b/fs/ocfs2/blockcheck.c
index 1d5083cef3a..f102ec939c9 100644
--- a/fs/ocfs2/blockcheck.c
+++ b/fs/ocfs2/blockcheck.c
@@ -39,6 +39,35 @@
* c = # total code bits (d + p)
*/
+
+/*
+ * Find the log base 2 of 32-bit v.
+ *
+ * Algorithm found on http://graphics.stanford.edu/~seander/bithacks.html,
+ * by Sean Eron Anderson. Code on the page is in the public domain unless
+ * otherwise noted.
+ *
+ * This particular algorithm is credited to Eric Cole.
+ */
+static int find_highest_bit_set(unsigned int v)
+{
+
+ static const int MultiplyDeBruijnBitPosition[32] =
+ {
+ 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
+ 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
+ };
+
+ v |= v >> 1; /* first round down to power of 2 */
+ v |= v >> 2;
+ v |= v >> 4;
+ v |= v >> 8;
+ v |= v >> 16;
+ v = (v >> 1) + 1;
+
+ return MultiplyDeBruijnBitPosition[(u32)(v * 0x077CB531UL) >> 27];
+}
+
/*
* Calculate the bit offset in the hamming code buffer based on the bit's
* offset in the data buffer. Since the hamming code reserves all
@@ -64,12 +93,21 @@ static unsigned int calc_code_bit(unsigned int i)
b = i + 1;
/*
+ * As a cheat, we know that all bits below b's highest bit must be
+ * parity bits, so we can start there.
+ */
+ p = find_highest_bit_set(b);
+ b += p;
+
+ /*
* For every power of two below our bit number, bump our bit.
*
* We compare with (b + 1) becuase we have to compare with what b
* would be _if_ it were bumped up by the parity bit. Capice?
+ *
+ * We start p at 2^p because of the cheat above.
*/
- for (p = 0; (1 << p) < (b + 1); p++)
+ for (p = (1 << p); p < (b + 1); p <<= 1)
b++;
return b;