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Commit 7bb458a5 authored by Joel Becker's avatar Joel Becker Committed by Mark Fasheh
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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: default avatarJoel Becker <joel.becker@oracle.com>
Signed-off-by: default avatarMark Fasheh <mfasheh@suse.com>
parent e798b3f8

fs/ocfs2/blockcheck.c

+39 −1
Original line number Diff line number Diff line
@@ -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;