On Wed, Apr 30, 2014 at 4:43 PM, Al Viro <v...@zeniv.linux.org.uk> wrote:
>
> OK, done and force-pushed. Should propagate in a few...
That made it more obvious how the DCACHE_MAY_FREE case ends up
working. And in particular, mind rewriting this:
if (dentry->d_flags & DCACHE_MAY_FREE) {
spin_unlock(&dentry->d_lock);
dentry_free(dentry);
} else {
spin_unlock(&dentry->d_lock);
}
return parent;
as just
bool free = dentry->d_flags & DCACHE_MAY_FREE;
spin_unlock(&dentry->d_lock);
if (free)
dentry_free(dentry);
return parent;
instead? In fact, I get the feeling that the other case later on
really fits the same model:
spin_lock(&dentry->d_lock);
if (dentry->d_flags & DCACHE_SHRINK_LIST) {
dentry->d_flags |= DCACHE_MAY_FREE;
spin_unlock(&dentry->d_lock);
} else {
spin_unlock(&dentry->d_lock);
dentry_free(dentry);
}
ends up really being better as
spin_lock(&dentry->d_lock);
free = 1;
if (dentry->d_flags & DCACHE_SHRINK_LIST) {
dentry->d_flags |= DCACHE_MAY_FREE;
free = 0;
}
spin_unlock(&dentry->d_lock);
if (free)
dentry_free(dentry);
return parent;
and then suddenly it looks like we have a common exit sequence from
that dentry_kill() function, no?
(The earlier "unlock_on_failure" exit case is altogether a different case).
I dunno. Maybe not a big deal, but one reason I prefer doing that
"free" flag is because I really tend to prefer the simple case of
lock-unlock pairing cleanly at the same level. NOT the pattern where
you have one lock at one indentation level, paired with multiple
unlocks for all the different cases.
Linus
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