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fs/locks: print full locks information
Commit fd7732e033
("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: Luo Longjun <luolongjun@huawei.com>
Signed-off-by: Jeff Layton <jlayton@kernel.org>
This commit is contained in:
parent
a38fd87484
commit
b8da9b10e2
65
fs/locks.c
65
fs/locks.c
@ -2828,7 +2828,7 @@ struct locks_iterator {
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};
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static void lock_get_status(struct seq_file *f, struct file_lock *fl,
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loff_t id, char *pfx)
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loff_t id, char *pfx, int repeat)
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{
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struct inode *inode = NULL;
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unsigned int fl_pid;
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@ -2844,7 +2844,11 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl,
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if (fl->fl_file != NULL)
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inode = locks_inode(fl->fl_file);
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seq_printf(f, "%lld:%s ", id, pfx);
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seq_printf(f, "%lld: ", id);
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if (repeat)
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seq_printf(f, "%*s", repeat - 1 + (int)strlen(pfx), pfx);
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if (IS_POSIX(fl)) {
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if (fl->fl_flags & FL_ACCESS)
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seq_puts(f, "ACCESS");
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@ -2906,21 +2910,64 @@ static void lock_get_status(struct seq_file *f, struct file_lock *fl,
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}
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}
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static struct file_lock *get_next_blocked_member(struct file_lock *node)
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{
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struct file_lock *tmp;
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/* NULL node or root node */
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if (node == NULL || node->fl_blocker == NULL)
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return NULL;
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/* Next member in the linked list could be itself */
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tmp = list_next_entry(node, fl_blocked_member);
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if (list_entry_is_head(tmp, &node->fl_blocker->fl_blocked_requests, fl_blocked_member)
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|| tmp == node) {
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return NULL;
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}
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return tmp;
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}
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static int locks_show(struct seq_file *f, void *v)
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{
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struct locks_iterator *iter = f->private;
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struct file_lock *fl, *bfl;
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struct file_lock *cur, *tmp;
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struct pid_namespace *proc_pidns = proc_pid_ns(file_inode(f->file)->i_sb);
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int level = 0;
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fl = hlist_entry(v, struct file_lock, fl_link);
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cur = hlist_entry(v, struct file_lock, fl_link);
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if (locks_translate_pid(fl, proc_pidns) == 0)
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if (locks_translate_pid(cur, proc_pidns) == 0)
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return 0;
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lock_get_status(f, fl, iter->li_pos, "");
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/* View this crossed linked list as a binary tree, the first member of fl_blocked_requests
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* is the left child of current node, the next silibing in fl_blocked_member is the
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* right child, we can alse get the parent of current node from fl_blocker, so this
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* question becomes traversal of a binary tree
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*/
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while (cur != NULL) {
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if (level)
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lock_get_status(f, cur, iter->li_pos, "-> ", level);
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else
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lock_get_status(f, cur, iter->li_pos, "", level);
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list_for_each_entry(bfl, &fl->fl_blocked_requests, fl_blocked_member)
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lock_get_status(f, bfl, iter->li_pos, " ->");
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if (!list_empty(&cur->fl_blocked_requests)) {
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/* Turn left */
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cur = list_first_entry_or_null(&cur->fl_blocked_requests,
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struct file_lock, fl_blocked_member);
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level++;
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} else {
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/* Turn right */
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tmp = get_next_blocked_member(cur);
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/* Fall back to parent node */
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while (tmp == NULL && cur->fl_blocker != NULL) {
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cur = cur->fl_blocker;
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level--;
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tmp = get_next_blocked_member(cur);
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}
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cur = tmp;
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}
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}
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return 0;
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}
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@ -2941,7 +2988,7 @@ static void __show_fd_locks(struct seq_file *f,
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(*id)++;
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seq_puts(f, "lock:\t");
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lock_get_status(f, fl, *id, "");
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lock_get_status(f, fl, *id, "", 0);
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}
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}
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