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btrfs: simplify extracting delayed node at btrfs_first_delayed_node()

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Instead of grabbing the next pointer from the list and then doing a
list_entry() call, we can simply use list_first_entry(), removing the need
for list_head variable.

Also there's no need to check if the list is empty before attempting to
extract the first element, we can use list_first_entry_or_null(), removing
the need for a special if statement and the 'out' label.

Reviewed-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
This commit is contained in:
Filipe Manana
2025-05-01 13:05:10 +01:00
committed by David Sterba
parent c5d12d5b62
commit 32bc875cbc
1 changed files with 5 additions and 9 deletions
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14
fs/btrfs/delayed-inode.c
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@@ -216,17 +216,13 @@ static void btrfs_dequeue_delayed_node(struct btrfs_delayed_root *root,
static struct btrfs_delayed_node *btrfs_first_delayed_node(
struct btrfs_delayed_root *delayed_root)
{
struct list_head *p;
struct btrfs_delayed_node *node = NULL;
struct btrfs_delayed_node *node;
spin_lock(&delayed_root->lock);
if (list_empty(&delayed_root->node_list))
goto out;
p = delayed_root->node_list.next;
node = list_entry(p, struct btrfs_delayed_node, n_list);
refcount_inc(&node->refs);
out:
node = list_first_entry_or_null(&delayed_root->node_list,
struct btrfs_delayed_node, n_list);
if (node)
refcount_inc(&node->refs);
spin_unlock(&delayed_root->lock);
return node;