vfs: factor out lookup_mountpoint from new_mountpoint
I am shortly going to add a new user of struct mountpoint that
needs to look up existing entries but does not want to create
a struct mountpoint if one does not exist. Therefore to keep
the code simple and easy to read split out lookup_mountpoint
from new_mountpoint.
Signed-off-by: "Eric W. Biederman" <ebiederm@xmission.com>
Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>
diff --git a/fs/namespace.c b/fs/namespace.c
index 99572dd..88fc3f4 100644
--- a/fs/namespace.c
+++ b/fs/namespace.c
@@ -703,11 +703,10 @@
return is_covered;
}
-static struct mountpoint *new_mountpoint(struct dentry *dentry)
+static struct mountpoint *lookup_mountpoint(struct dentry *dentry)
{
struct hlist_head *chain = mp_hash(dentry);
struct mountpoint *mp;
- int ret;
hlist_for_each_entry(mp, chain, m_hash) {
if (mp->m_dentry == dentry) {
@@ -718,6 +717,14 @@
return mp;
}
}
+ return NULL;
+}
+
+static struct mountpoint *new_mountpoint(struct dentry *dentry)
+{
+ struct hlist_head *chain = mp_hash(dentry);
+ struct mountpoint *mp;
+ int ret;
mp = kmalloc(sizeof(struct mountpoint), GFP_KERNEL);
if (!mp)
@@ -1818,7 +1825,9 @@
namespace_lock();
mnt = lookup_mnt(path);
if (likely(!mnt)) {
- struct mountpoint *mp = new_mountpoint(dentry);
+ struct mountpoint *mp = lookup_mountpoint(dentry);
+ if (!mp)
+ mp = new_mountpoint(dentry);
if (IS_ERR(mp)) {
namespace_unlock();
mutex_unlock(&dentry->d_inode->i_mutex);