aboutsummaryrefslogtreecommitdiffstats
path: root/fs/ocfs2/blockcheck.c
diff options
context:
space:
mode:
Diffstat (limited to 'fs/ocfs2/blockcheck.c')
-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 1d5083cef3a2..f102ec939c90 100644
--- a/fs/ocfs2/blockcheck.c
+++ b/fs/ocfs2/blockcheck.c
@@ -39,6 +39,35 @@
39 * c = # total code bits (d + p) 39 * c = # total code bits (d + p)
40 */ 40 */
41 41
42
43/*
44 * Find the log base 2 of 32-bit v.
45 *
46 * Algorithm found on http://graphics.stanford.edu/~seander/bithacks.html,
47 * by Sean Eron Anderson. Code on the page is in the public domain unless
48 * otherwise noted.
49 *
50 * This particular algorithm is credited to Eric Cole.
51 */
52static int find_highest_bit_set(unsigned int v)
53{
54
55 static const int MultiplyDeBruijnBitPosition[32] =
56 {
57 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
58 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
59 };
60
61 v |= v >> 1; /* first round down to power of 2 */
62 v |= v >> 2;
63 v |= v >> 4;
64 v |= v >> 8;
65 v |= v >> 16;
66 v = (v >> 1) + 1;
67
68 return MultiplyDeBruijnBitPosition[(u32)(v * 0x077CB531UL) >> 27];
69}
70
42/* 71/*
43 * Calculate the bit offset in the hamming code buffer based on the bit's 72 * Calculate the bit offset in the hamming code buffer based on the bit's
44 * offset in the data buffer. Since the hamming code reserves all 73 * offset in the data buffer. Since the hamming code reserves all
@@ -64,12 +93,21 @@ static unsigned int calc_code_bit(unsigned int i)
64 b = i + 1; 93 b = i + 1;
65 94
66 /* 95 /*
96 * As a cheat, we know that all bits below b's highest bit must be
97 * parity bits, so we can start there.
98 */
99 p = find_highest_bit_set(b);
100 b += p;
101
102 /*
67 * For every power of two below our bit number, bump our bit. 103 * For every power of two below our bit number, bump our bit.
68 * 104 *
69 * We compare with (b + 1) becuase we have to compare with what b 105 * We compare with (b + 1) becuase we have to compare with what b
70 * would be _if_ it were bumped up by the parity bit. Capice? 106 * would be _if_ it were bumped up by the parity bit. Capice?
107 *
108 * We start p at 2^p because of the cheat above.
71 */ 109 */
72 for (p = 0; (1 << p) < (b + 1); p++) 110 for (p = (1 << p); p < (b + 1); p <<= 1)
73 b++; 111 b++;
74 112
75 return b; 113 return b;