proc: actually make proc_fd_permission() thread-friendly
The commit96d0df79f2
("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). Fixes:96d0df79f2
("proc: make proc_fd_permission() thread-friendly") Reported-by: "Jin, Yihua" <yihua.jin@intel.com> Signed-off-by: Oleg Nesterov <oleg@redhat.com> Cc: "Eric W. Biederman" <ebiederm@xmission.com> Cc: <stable@vger.kernel.org> Signed-off-by: Andrew Morton <akpm@linux-foundation.org> Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
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3a49f3d2a1
commit
54708d2858
14
fs/proc/fd.c
14
fs/proc/fd.c
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@ -291,11 +291,19 @@ static struct dentry *proc_lookupfd(struct inode *dir, struct dentry *dentry,
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*/
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int proc_fd_permission(struct inode *inode, int mask)
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{
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int rv = generic_permission(inode, mask);
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struct task_struct *p;
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int rv;
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rv = generic_permission(inode, mask);
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if (rv == 0)
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return 0;
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if (task_tgid(current) == proc_pid(inode))
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return rv;
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rcu_read_lock();
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p = pid_task(proc_pid(inode), PIDTYPE_PID);
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if (p && same_thread_group(p, current))
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rv = 0;
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rcu_read_unlock();
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return rv;
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}
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