On Tue, Aug 8, 2017 at 10:13 PM, Gao Feng <gfree.w...@vip.163.com> wrote: > Maybe I didn't show my explanation clearly. > I think it won't happen as I mentioned in the last email. > Because the pptp_release invokes the synchronize_rcu to make sure it, and > actually there is no one which would invoke del_chan except pptp_release. > It is guaranteed by that the pptp_release doesn't put the sock refcnt until > complete all cleanup include marking sk_state as PPPOX_DEAD. > > In other words, even though the pptp_release is not the last user of this > sock, the other one wouldn't invoke del_chan in pptp_sock_destruct. > Because the condition "!(sk->sk_state & PPPOX_DEAD)" must be false.
Only if sock->sk is always non-NULL for pptp_release(), which is what I am not sure. If you look at other ->release(), similar checks are there too, so not just for pptp. > > As summary, the del_chan and pppox_unbind_sock in pptp_sock_destruct are > unnecessary. > And it even brings confusing. Sorry, I can't draw any conclusion for this.
