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ocfs2: Another hamming code optimization.

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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>
This commit is contained in:
Joel Becker 2008-12-15 18:24:33 -08:00 committed by Mark Fasheh
parent e798b3f8a9
commit 7bb458a585
1 changed files with 39 additions and 1 deletions
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40
fs/ocfs2/blockcheck.c
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@ -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;