[PATCH] proc: actually make proc_fd_permission() thread-friendly

From: Oleg Nesterov
Date: Tue Nov 03 2015 - 11:42:42 EST


The commit 96d0df79f264 ("proc: make proc_fd_permission() thread-friendly")
fixed the access to /proc/self/fd from sub-threads, but introduced another
problem: a sub-thread can't access /proc/<tid>/fd/ or /proc/thread-self/fd
if generic_permission() fails.

Change proc_fd_permission() to check same_thread_group(pid_task(), current).

Reported-by: "Jin, Yihua" <yihua.jin@xxxxxxxxx>
Signed-off-by: Oleg Nesterov <oleg@xxxxxxxxxx>
Cc: stable@xxxxxxxxxxxxxxx
---
fs/proc/fd.c | 14 +++++++++++---
1 files changed, 11 insertions(+), 3 deletions(-)

diff --git a/fs/proc/fd.c b/fs/proc/fd.c
index 6e5fcd0..3c2a915 100644
--- a/fs/proc/fd.c
+++ b/fs/proc/fd.c
@@ -291,11 +291,19 @@ static struct dentry *proc_lookupfd(struct inode *dir, struct dentry *dentry,
*/
int proc_fd_permission(struct inode *inode, int mask)
{
- int rv = generic_permission(inode, mask);
+ struct task_struct *p;
+ int rv;
+
+ rv = generic_permission(inode, mask);
if (rv == 0)
- return 0;
- if (task_tgid(current) == proc_pid(inode))
+ return rv;
+
+ rcu_read_lock();
+ p = pid_task(proc_pid(inode), PIDTYPE_PID);
+ if (p && same_thread_group(p, current))
rv = 0;
+ rcu_read_unlock();
+
return rv;
}

--
1.5.5.1


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