Commit b8da9b10 authored by Luo Longjun's avatar Luo Longjun Committed by Jeff Layton

fs/locks: print full locks information

Commit fd7732e0 ("fs/locks: create a tree of dependent requests.")
has put blocked locks into a tree.

So, with a for loop, we can't check all locks information.

To solve this problem, we should traverse the tree.
Signed-off-by: default avatarLuo Longjun <luolongjun@huawei.com>
Signed-off-by: default avatarJeff Layton <jlayton@kernel.org>
parent a38fd874
...@@ -2828,7 +2828,7 @@ struct locks_iterator { ...@@ -2828,7 +2828,7 @@ struct locks_iterator {
}; };
static void lock_get_status(struct seq_file *f, struct file_lock *fl, static void lock_get_status(struct seq_file *f, struct file_lock *fl,
loff_t id, char *pfx) loff_t id, char *pfx, int repeat)
{ {
struct inode *inode = NULL; struct inode *inode = NULL;
unsigned int fl_pid; unsigned int fl_pid;
...@@ -2844,7 +2844,11 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl, ...@@ -2844,7 +2844,11 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl,
if (fl->fl_file != NULL) if (fl->fl_file != NULL)
inode = locks_inode(fl->fl_file); inode = locks_inode(fl->fl_file);
seq_printf(f, "%lld:%s ", id, pfx); seq_printf(f, "%lld: ", id);
if (repeat)
seq_printf(f, "%*s", repeat - 1 + (int)strlen(pfx), pfx);
if (IS_POSIX(fl)) { if (IS_POSIX(fl)) {
if (fl->fl_flags & FL_ACCESS) if (fl->fl_flags & FL_ACCESS)
seq_puts(f, "ACCESS"); seq_puts(f, "ACCESS");
...@@ -2906,21 +2910,64 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl, ...@@ -2906,21 +2910,64 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl,
} }
} }
static struct file_lock *get_next_blocked_member(struct file_lock *node)
{
struct file_lock *tmp;
/* NULL node or root node */
if (node == NULL || node->fl_blocker == NULL)
return NULL;
/* Next member in the linked list could be itself */
tmp = list_next_entry(node, fl_blocked_member);
if (list_entry_is_head(tmp, &node->fl_blocker->fl_blocked_requests, fl_blocked_member)
|| tmp == node) {
return NULL;
}
return tmp;
}
static int locks_show(struct seq_file *f, void *v) static int locks_show(struct seq_file *f, void *v)
{ {
struct locks_iterator *iter = f->private; struct locks_iterator *iter = f->private;
struct file_lock *fl, *bfl; struct file_lock *cur, *tmp;
struct pid_namespace *proc_pidns = proc_pid_ns(file_inode(f->file)->i_sb); struct pid_namespace *proc_pidns = proc_pid_ns(file_inode(f->file)->i_sb);
int level = 0;
fl = hlist_entry(v, struct file_lock, fl_link); cur = hlist_entry(v, struct file_lock, fl_link);
if (locks_translate_pid(fl, proc_pidns) == 0) if (locks_translate_pid(cur, proc_pidns) == 0)
return 0; return 0;
lock_get_status(f, fl, iter->li_pos, ""); /* View this crossed linked list as a binary tree, the first member of fl_blocked_requests
* is the left child of current node, the next silibing in fl_blocked_member is the
* right child, we can alse get the parent of current node from fl_blocker, so this
* question becomes traversal of a binary tree
*/
while (cur != NULL) {
if (level)
lock_get_status(f, cur, iter->li_pos, "-> ", level);
else
lock_get_status(f, cur, iter->li_pos, "", level);
list_for_each_entry(bfl, &fl->fl_blocked_requests, fl_blocked_member) if (!list_empty(&cur->fl_blocked_requests)) {
lock_get_status(f, bfl, iter->li_pos, " ->"); /* Turn left */
cur = list_first_entry_or_null(&cur->fl_blocked_requests,
struct file_lock, fl_blocked_member);
level++;
} else {
/* Turn right */
tmp = get_next_blocked_member(cur);
/* Fall back to parent node */
while (tmp == NULL && cur->fl_blocker != NULL) {
cur = cur->fl_blocker;
level--;
tmp = get_next_blocked_member(cur);
}
cur = tmp;
}
}
return 0; return 0;
} }
...@@ -2941,7 +2988,7 @@ static void __show_fd_locks(struct seq_file *f, ...@@ -2941,7 +2988,7 @@ static void __show_fd_locks(struct seq_file *f,
(*id)++; (*id)++;
seq_puts(f, "lock:\t"); seq_puts(f, "lock:\t");
lock_get_status(f, fl, *id, ""); lock_get_status(f, fl, *id, "", 0);
} }
} }
......
Markdown is supported
0%
or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment