On Tue, Feb 20, 2018 at 12:34:57AM +0100, John Ogness wrote:

> Implementation 3: The same as implementation 2 but using if's to
> support branch prediction. This approach is probably the most
> complicated to understand but will be the fastest.
> 
>               /*
>                * Lock the inode. Might drop dentry->d_lock temporarily
>                * which allows inode to change. Start over if that happens.
>                */
>                 int ret = dentry_lock_inode(dentry);
>                 if (unlikely(ret != LOCK_FAST)) {
>                         if (ret == LOCK_FAILED)
>                                 goto again;
>                         /*
>                          * Recheck refcount as it might have been
>                          * incremented while d_lock was dropped.
>                          */
>                         if (dentry->d_lockref.count != 1)
>                                 goto drop_ref;
>                 }

Implementation 4: screw the tristate, move the loop inside dentry_lock_inode().

> If lock_parent() returns a non-NULL, it is returning
> dentry->d_parent. So the return value is really just a boolean and the
> locked parent is the parent of the dentry. The function is a little bit
> tricky because it could return NULL (lock failed) even if the dentry has
> a non-NULL d_parent. So any caller using a tristate return variation of
> lock_parent() must rely on the return value instead of a non-NULL
> dentry->d_parent.

dentry always has non-NULL ->d_parent; it might point to dentry itself, but
it's never NULL.

>               if (!dentry->d_lockref.count) {
> -                     struct dentry *parent = lock_parent(dentry);
> +                     int ret = lock_parent(dentry);
> +                     parent = dentry->d_parent;
>                       if (likely(!dentry->d_lockref.count)) {
>                               __dentry_kill(dentry);
>                               dput(parent);

Broken.  In IS_ROOT case you'll hit an extra dput() on dentry itself.
dput(NULL) is no-op; this, OTOH, isn't.

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