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>
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@ -39,6 +39,35 @@
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* c = # total code bits (d + p)
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* c = # total code bits (d + p)
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*/
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*/
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/*
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* Find the log base 2 of 32-bit v.
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*
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* Algorithm found on http://graphics.stanford.edu/~seander/bithacks.html,
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* by Sean Eron Anderson. Code on the page is in the public domain unless
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* otherwise noted.
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*
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* This particular algorithm is credited to Eric Cole.
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*/
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static int find_highest_bit_set(unsigned int v)
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{
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static const int MultiplyDeBruijnBitPosition[32] =
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{
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0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
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31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
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};
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v |= v >> 1; /* first round down to power of 2 */
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v |= v >> 2;
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v |= v >> 4;
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v |= v >> 8;
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v |= v >> 16;
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v = (v >> 1) + 1;
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return MultiplyDeBruijnBitPosition[(u32)(v * 0x077CB531UL) >> 27];
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}
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/*
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/*
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* Calculate the bit offset in the hamming code buffer based on the bit's
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* Calculate the bit offset in the hamming code buffer based on the bit's
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* offset in the data buffer. Since the hamming code reserves all
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* offset in the data buffer. Since the hamming code reserves all
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@ -63,13 +92,22 @@ static unsigned int calc_code_bit(unsigned int i)
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*/
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*/
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b = i + 1;
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b = i + 1;
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/*
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* As a cheat, we know that all bits below b's highest bit must be
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* parity bits, so we can start there.
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*/
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p = find_highest_bit_set(b);
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b += p;
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/*
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/*
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* For every power of two below our bit number, bump our bit.
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* For every power of two below our bit number, bump our bit.
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*
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*
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* We compare with (b + 1) becuase we have to compare with what b
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* We compare with (b + 1) becuase we have to compare with what b
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* would be _if_ it were bumped up by the parity bit. Capice?
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* would be _if_ it were bumped up by the parity bit. Capice?
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*
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* We start p at 2^p because of the cheat above.
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*/
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*/
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for (p = 0; (1 << p) < (b + 1); p++)
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for (p = (1 << p); p < (b + 1); p <<= 1)
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b++;
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b++;
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return b;
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return b;
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