Date: Thu, 20 Mar 2025 10:16:58 +0000 From: bugzilla-noreply@freebsd.org To: net@FreeBSD.org Subject: [Bug 283426] panic in sbappendaddr_locked() - if_ovpn related? Message-ID: <bug-283426-7501-DkjLh4haxD@https.bugs.freebsd.org/bugzilla/> In-Reply-To: <bug-283426-7501@https.bugs.freebsd.org/bugzilla/> References: <bug-283426-7501@https.bugs.freebsd.org/bugzilla/>
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https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=3D283426 Bjoern A. Zeeb <bz@FreeBSD.org> changed: What |Removed |Added ---------------------------------------------------------------------------- Assignee|wireless@FreeBSD.org |net@FreeBSD.org Summary|panic in |panic in |sbappendaddr_locked() |sbappendaddr_locked() - | |if_ovpn related? CC| |kp@freebsd.org --- Comment #12 from Bjoern A. Zeeb <bz@FreeBSD.org> --- (In reply to Robin Haberkorn from comment #11) Yes I understand why the question was for the original one; I just wanted = to understand as the "copy" should have it too then and when manually going up= the stack I couldn't spot where it might be cleared. I wonder what Dan Epure's report looked like, Glebius? Do you have a backt= race from that? Same code paths? Also UDP? Also ovpn loaded? I don't know how the openvpn kernel module is hooking into the input path; = if it's using u_tun_func then that at least would explain why we do not see it= in the backtrace. And then Takahiro Kurosawa's analysis from #c7 sounds fairly valid if m_unshare can lose the PKTHDR on the copy. OpenVPN doesn't have t= o be active; it seems if there's an interface the filtering function will be ho= oked up. And then it's not a regression in iwm or the network stack; then it's imply the feature. Adding Kristof and bouncing this back to net@. --=20 You are receiving this mail because: You are the assignee for the bug.=
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