Commit 4ec331c3 authored by Joe Thornber's avatar Joe Thornber Committed by Mike Snitzer

dm btree: add dm_btree_remove_leaves()

Removes a range of leaf values from the tree.
Signed-off-by: default avatarJoe Thornber <ejt@redhat.com>
Signed-off-by: default avatarMike Snitzer <snitzer@redhat.com>
parent 0f24b79b
...@@ -590,3 +590,130 @@ int dm_btree_remove(struct dm_btree_info *info, dm_block_t root, ...@@ -590,3 +590,130 @@ int dm_btree_remove(struct dm_btree_info *info, dm_block_t root,
return r; return r;
} }
EXPORT_SYMBOL_GPL(dm_btree_remove); EXPORT_SYMBOL_GPL(dm_btree_remove);
/*----------------------------------------------------------------*/
static int remove_nearest(struct shadow_spine *s, struct dm_btree_info *info,
struct dm_btree_value_type *vt, dm_block_t root,
uint64_t key, int *index)
{
int i = *index, r;
struct btree_node *n;
for (;;) {
r = shadow_step(s, root, vt);
if (r < 0)
break;
/*
* We have to patch up the parent node, ugly, but I don't
* see a way to do this automatically as part of the spine
* op.
*/
if (shadow_has_parent(s)) {
__le64 location = cpu_to_le64(dm_block_location(shadow_current(s)));
memcpy(value_ptr(dm_block_data(shadow_parent(s)), i),
&location, sizeof(__le64));
}
n = dm_block_data(shadow_current(s));
if (le32_to_cpu(n->header.flags) & LEAF_NODE) {
*index = lower_bound(n, key);
return 0;
}
r = rebalance_children(s, info, vt, key);
if (r)
break;
n = dm_block_data(shadow_current(s));
if (le32_to_cpu(n->header.flags) & LEAF_NODE) {
*index = lower_bound(n, key);
return 0;
}
i = lower_bound(n, key);
/*
* We know the key is present, or else
* rebalance_children would have returned
* -ENODATA
*/
root = value64(n, i);
}
return r;
}
static int remove_one(struct dm_btree_info *info, dm_block_t root,
uint64_t *keys, uint64_t end_key,
dm_block_t *new_root, unsigned *nr_removed)
{
unsigned level, last_level = info->levels - 1;
int index = 0, r = 0;
struct shadow_spine spine;
struct btree_node *n;
uint64_t k;
init_shadow_spine(&spine, info);
for (level = 0; level < last_level; level++) {
r = remove_raw(&spine, info, &le64_type,
root, keys[level], (unsigned *) &index);
if (r < 0)
goto out;
n = dm_block_data(shadow_current(&spine));
root = value64(n, index);
}
r = remove_nearest(&spine, info, &info->value_type,
root, keys[last_level], &index);
if (r < 0)
goto out;
n = dm_block_data(shadow_current(&spine));
if (index < 0)
index = 0;
if (index >= le32_to_cpu(n->header.nr_entries)) {
r = -ENODATA;
goto out;
}
k = le64_to_cpu(n->keys[index]);
if (k >= keys[last_level] && k < end_key) {
if (info->value_type.dec)
info->value_type.dec(info->value_type.context,
value_ptr(n, index));
delete_at(n, index);
} else
r = -ENODATA;
out:
*new_root = shadow_root(&spine);
exit_shadow_spine(&spine);
return r;
}
int dm_btree_remove_leaves(struct dm_btree_info *info, dm_block_t root,
uint64_t *first_key, uint64_t end_key,
dm_block_t *new_root, unsigned *nr_removed)
{
int r;
*nr_removed = 0;
do {
r = remove_one(info, root, first_key, end_key, &root, nr_removed);
if (!r)
(*nr_removed)++;
} while (!r);
*new_root = root;
return r == -ENODATA ? 0 : r;
}
EXPORT_SYMBOL_GPL(dm_btree_remove_leaves);
...@@ -134,6 +134,15 @@ int dm_btree_insert_notify(struct dm_btree_info *info, dm_block_t root, ...@@ -134,6 +134,15 @@ int dm_btree_insert_notify(struct dm_btree_info *info, dm_block_t root,
int dm_btree_remove(struct dm_btree_info *info, dm_block_t root, int dm_btree_remove(struct dm_btree_info *info, dm_block_t root,
uint64_t *keys, dm_block_t *new_root); uint64_t *keys, dm_block_t *new_root);
/*
* Removes values between 'keys' and keys2, where keys2 is keys with the
* final key replaced with 'end_key'. 'end_key' is the one-past-the-end
* value. 'keys' may be altered.
*/
int dm_btree_remove_leaves(struct dm_btree_info *info, dm_block_t root,
uint64_t *keys, uint64_t end_key,
dm_block_t *new_root, unsigned *nr_removed);
/* /*
* Returns < 0 on failure. Otherwise the number of key entries that have * Returns < 0 on failure. Otherwise the number of key entries that have
* been filled out. Remember trees can have zero entries, and as such have * been filled out. Remember trees can have zero entries, and as such have
......
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