The above is roughly how the current code tries to work, except it's
broken and assumes the number of dentries to be 1.
However, the problem with the code is that a dentry now depends on two
things: both iput() and dput() can result in a dentry being free. And that
is why the code is so complex: dput cannot on its own decide whether it
can free stuff, it needs help from iput..
Generally, I don't want those kinds of dependencies: at least in my
opinion a circular dependency (inodes depend on dentries which depend on
inodes) tends to drive complexity not up by two, but by a power of two..
Anyway, that's why I've been re-doing the logic so that dentries no longer
depend on any inode information: when the dentry count goes down to zero
we can always just free the dentry regardless of any inode counts
(essentially because the inode count has now been relegated to be just a
"dentry link count").
Linus