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namei: make set_root_rcu() return void

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The only caller that cares about its return value can just
as easily pick it from nd->root_seq itself.  We used to just
calculate it and return to caller, but these days we are
storing it in nd->root_seq in all cases.

Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>
This commit is contained in:
Al Viro 2015-06-29 12:07:04 -04:00
parent dc3f4198ea
commit 06d7137e5c
1 changed files with 3 additions and 3 deletions
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6
fs/namei.c
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@ -792,7 +792,7 @@ static void set_root(struct nameidata *nd)
get_fs_root(current->fs, &nd->root);
}
static unsigned set_root_rcu(struct nameidata *nd)
static void set_root_rcu(struct nameidata *nd)
{
struct fs_struct *fs = current->fs;
unsigned seq;
@ -802,7 +802,6 @@ static unsigned set_root_rcu(struct nameidata *nd)
nd->root = fs->root;
nd->root_seq = __read_seqcount_begin(&nd->root.dentry->d_seq);
} while (read_seqcount_retry(&fs->seq, seq));
return nd->root_seq;
}
static void path_put_conditional(struct path *path, struct nameidata *nd)
@ -1998,7 +1997,8 @@ static const char *path_init(struct nameidata *nd, unsigned flags)
if (*s == '/') {
if (flags & LOOKUP_RCU) {
rcu_read_lock();
nd->seq = set_root_rcu(nd);
set_root_rcu(nd);
nd->seq = nd->root_seq;
} else {
set_root(nd);
path_get(&nd->root);