Date: Tue, 22 Feb 2005 10:36:18 -0500 From: Brian Fundakowski Feldman <green@freebsd.org> To: Divacky Roman <xdivac02@stud.fit.vutbr.cz> Cc: current@freebsd.org Subject: Re: pfctl -f causes fatal trap 12 Message-ID: <20050222153618.GA966@green.homeunix.org> In-Reply-To: <20050213074020.GA28974@stud.fit.vutbr.cz> References: <20050212030852.GF693@dmr.ath.cx> <20050212162440.GA82987@stud.fit.vutbr.cz> <20050213005144.GA2580@dmr.ath.cx> <20050213074020.GA28974@stud.fit.vutbr.cz>
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On Sun, Feb 13, 2005 at 08:40:20AM +0100, Divacky Roman wrote: > On Sun, Feb 13, 2005 at 11:51:44AM +1100, Emil Mikulic wrote: > > On Sat, Feb 12, 2005, Divacky Roman top-posted: > > > same here.... I have pf as module and see this panic on boot > > > > > > when I reboot after the panic, fsck takes place and then it doesn > > > panic (obviously the fsck changes something which causes pf not to > > > panic) > > > > Really? I hadn't noticed that. You should send your message to the > > mailing list in case it can help one of the developers track down the > > problem. > > yes.... and oh, I forgot to cc: current (and I am doing it now) > > > >From the kgdb output: > > > > (kgdb) frame 11 > > > > #11 0xc0627fdc in softclock (dummy=0x0) at \ > > > > /usr/src/sys/kern/kern_timeout.c:315 > > > > 315 mtx_unlock(c_mtx); > > > > (kgdb) print c_mtx > > > > $1 = (struct mtx *) 0x0 > > > > It looks to me like that part of the code is trying to unlock an > > uninitialized mutex. > > > > I guess running fsck somehow causes the mutex to be initialized before > > loading the pf rules breaks stuff. This should be fixed now -- callout_init() was not being called. -- Brian Fundakowski Feldman \'[ FreeBSD ]''''''''''\ <> green@FreeBSD.org \ The Power to Serve! \ Opinions expressed are my own. \,,,,,,,,,,,,,,,,,,,,,,\
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