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scoutfs: refactor btree split condition

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Btree traversal doesn't split a block if it has room for the caller's
item.  Extract this test into a function so that an upcoming btree call
can test that each of multiple insertions into a leaf will fit.

Signed-off-by: Zach Brown <zab@versity.com>
This commit is contained in:
Zach Brown
2020-08-18 10:14:12 -07:00
committed by Zach Brown
parent d440056e6f
commit 9e975dffe1
1 changed files with 8 additions and 1 deletions
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9
kmod/src/btree.c
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@@ -153,6 +153,13 @@ static inline unsigned int mid_free_off(struct scoutfs_btree_block *bt)
return le16_to_cpu(ptr_off(bt, &bt->items[le16_to_cpu(bt->nr_items)]));
}
/* true if the mid free region has room for an item struct and its value */
static inline bool mid_free_item_room(struct scoutfs_btree_block *bt,
int val_len)
{
return le16_to_cpu(bt->mid_free_len) >= item_len_bytes(val_len);
}
static inline struct scoutfs_key *item_key(struct scoutfs_btree_item *item)
{
return &item->key;
@@ -873,7 +880,7 @@ static int try_split(struct super_block *sb,
val_len = sizeof(struct scoutfs_btree_ref);
/* don't need to split if there's enough space for the item */
if (le16_to_cpu(right->mid_free_len) >= item_len_bytes(val_len))
if (mid_free_item_room(right, val_len))
return 0;
if (item_full_pct(right) < 80) {