From 7bb458a58588f397068e4166c615e9fcc7480c16 Mon Sep 17 00:00:00 2001 From: Joel Becker Date: Mon, 15 Dec 2008 18:24:33 -0800 Subject: 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 Signed-off-by: Mark Fasheh --- fs/ocfs2/blockcheck.c | 40 +++++++++++++++++++++++++++++++++++++++- 1 file changed, 39 insertions(+), 1 deletion(-) 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 @@ -63,13 +92,22 @@ 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; -- cgit v1.2.3