btrfs: simplify some assignments of inode numbers
There are several places when the btrfs inode is converted to the generic inode, back to btrfs and then passed to btrfs_ino. We can remove the extra back and forth conversions. Signed-off-by: David Sterba <dsterba@suse.com>
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8f6c72a9e0
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3750851562
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@ -4242,9 +4242,9 @@ again:
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prev = node;
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entry = rb_entry(node, struct btrfs_inode, rb_node);
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if (objectid < btrfs_ino(BTRFS_I(&entry->vfs_inode)))
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if (objectid < btrfs_ino(entry))
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node = node->rb_left;
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else if (objectid > btrfs_ino(BTRFS_I(&entry->vfs_inode)))
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else if (objectid > btrfs_ino(entry))
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node = node->rb_right;
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else
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break;
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@ -4252,7 +4252,7 @@ again:
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if (!node) {
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while (prev) {
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entry = rb_entry(prev, struct btrfs_inode, rb_node);
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if (objectid <= btrfs_ino(BTRFS_I(&entry->vfs_inode))) {
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if (objectid <= btrfs_ino(entry)) {
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node = prev;
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break;
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}
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@ -4261,7 +4261,7 @@ again:
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}
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while (node) {
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entry = rb_entry(node, struct btrfs_inode, rb_node);
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objectid = btrfs_ino(BTRFS_I(&entry->vfs_inode)) + 1;
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objectid = btrfs_ino(entry) + 1;
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inode = igrab(&entry->vfs_inode);
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if (inode) {
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spin_unlock(&root->inode_lock);
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@ -5614,9 +5614,9 @@ static void inode_tree_add(struct inode *inode)
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parent = *p;
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entry = rb_entry(parent, struct btrfs_inode, rb_node);
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if (ino < btrfs_ino(BTRFS_I(&entry->vfs_inode)))
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if (ino < btrfs_ino(entry))
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p = &parent->rb_left;
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else if (ino > btrfs_ino(BTRFS_I(&entry->vfs_inode)))
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else if (ino > btrfs_ino(entry))
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p = &parent->rb_right;
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else {
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WARN_ON(!(entry->vfs_inode.i_state &
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