The current logic does not calculate correctly the good shift array:
Let x be the pattern that is being searched. Let y be the block of data.
The good shift array aligns the segment:
x[i+1 ... m-1] = y[i+j+1 ... j+m-1]
with its rightmost occurrence in x that fulfils x[i] neq y[i+j].
In previous version, the good shift array for the pattern ANPANMAN is:
[1, 8, 3, 8, 8, 8, 8, 8]
and should be:
[1, 8, 3, 6, 6, 6, 6, 6]
Signed-off-by: Pablo Neira Ayuso <pablo@netfilter.org>
Signed-off-by: David S. Miller <davem@davemloft.net>
+static int subpattern(u8 *pattern, int i, int j, int g)
+{
+ int x = i+g-1, y = j+g-1, ret = 0;
+
+ while(pattern[x--] == pattern[y--]) {
+ if (y < 0) {
+ ret = 1;
+ break;
+ }
+ if (--g == 0) {
+ ret = pattern[i-1] != pattern[j-1];
+ break;
+ }
+ }
+
+ return ret;
+}
+
static void compute_prefix_tbl(struct ts_bm *bm, const u8 *pattern,
unsigned int len)
{
static void compute_prefix_tbl(struct ts_bm *bm, const u8 *pattern,
unsigned int len)
{
- int i, j, ended, l[ASIZE];
for (i = 0; i < ASIZE; i++)
bm->bad_shift[i] = len;
for (i = 0; i < ASIZE; i++)
bm->bad_shift[i] = len;
/* Compute the good shift array, used to match reocurrences
* of a subpattern */
/* Compute the good shift array, used to match reocurrences
* of a subpattern */
- for (i = 1; i < bm->patlen; i++) {
- for (j = 0; j < bm->patlen && bm->pattern[bm->patlen - 1 - j]
- == bm->pattern[bm->patlen - 1 - i - j]; j++);
- l[i] = j;
- }
-
bm->good_shift[0] = 1;
for (i = 1; i < bm->patlen; i++)
bm->good_shift[i] = bm->patlen;
bm->good_shift[0] = 1;
for (i = 1; i < bm->patlen; i++)
bm->good_shift[i] = bm->patlen;
- for (i = bm->patlen - 1; i > 0; i--)
- bm->good_shift[l[i]] = i;
- ended = 0;
- for (i = 0; i < bm->patlen; i++) {
- if (l[i] == bm->patlen - 1 - i)
- ended = i;
- if (ended)
- bm->good_shift[i] = ended;
+ for (i = bm->patlen-1, g = 1; i > 0; g++, i--) {
+ for (j = i-1; j >= 1-g ; j--)
+ if (subpattern(bm->pattern, i, j, g)) {
+ bm->good_shift[g] = bm->patlen-j-g;
+ break;
+ }