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ldb_tdb: Rework list_union to not return duplicates, and keep sort order

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This allows the binary search to still operate on the list, even after
a or operator in the search expression

Signed-off-by: Andrew Bartlett <abartlet@samba.org>
Reviewed-by: Garming Sam <garming@catalyst.net.nz>
This commit is contained in:
Andrew Bartlett 2017-08-22 11:07:45 +12:00
parent b86a46df1d
commit fdff9a7087
1 changed files with 31 additions and 4 deletions
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35
lib/ldb/ldb_tdb/ldb_index.c
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@ -890,6 +890,7 @@ static bool list_union(struct ldb_context *ldb,
struct dn_list *list, struct dn_list *list2)
{
struct ldb_val *dn3;
unsigned int i = 0, j = 0, k = 0;
if (list2->count == 0) {
/* X | 0 == X */
@ -921,12 +922,38 @@ static bool list_union(struct ldb_context *ldb,
return false;
}
/* we allow for duplicates here, and get rid of them later */
memcpy(dn3, list->dn, sizeof(list->dn[0])*list->count);
memcpy(dn3+list->count, list2->dn, sizeof(list2->dn[0])*list2->count);
while (i < list->count || j < list2->count) {
int cmp;
if (i >= list->count) {
cmp = 1;
} else if (j >= list2->count) {
cmp = -1;
} else {
cmp = ldb_val_equal_exact_ordered(list->dn[i],
&list2->dn[j]);
}
if (cmp < 0) {
/* Take list */
dn3[k] = list->dn[i];
i++;
k++;
} else if (cmp > 0) {
/* Take list2 */
dn3[k] = list2->dn[j];
j++;
k++;
} else {
/* Equal, take list */
dn3[k] = list->dn[i];
i++;
j++;
k++;
}
}
list->dn = dn3;
list->count += list2->count;
list->count = k;
return true;
}