At 2017-08-10 02:18:44, "Cong Wang" <xiyou.wangc...@gmail.com> wrote: >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.
Yes. It seems only if the release() is invoked twice, the sock->sk would be NULL. But I don't find there is any case which could cause it. > >> >> 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. Thank you all the same, and I have learn a lot from you :) Wish someone which is familiar with these codes could give more details and explanations. Best Regards Feng
