On Wed, 19 Nov 2008, Tzafrir Cohen wrote:
> On Wed, Nov 19, 2008 at 07:57:33PM +0000, Jeff LaCoursiere wrote: >> >> ast% ps auxw | grep modprobe >> root 17744 99.9 0.0 2688 412 ? RN Nov03 23223:01 modprobe >> -r ipt_state > > modprobe -r is basically rmmod . rmmod and insmod and nowdays mostly > wrappers to kernel code. > > So while an strace of that process might give some more information > about it, I believe that the kernel-level backtrace would be more > interesting. > > For that, try either the 'p' or 't' sysrq commands. 'p' gives a stack > trace of the current process. 't': of all the processes. You can give a > sysrq command either through the console (on x86: alt-sysrq-<key>) or: > > echo <key> >/proc/sysrq-trigger No access to the console, sadly, so I tried the trigger method: [EMAIL PROTECTED] init.d]# echo p > /proc/sysrq-trigger which resulted in a single line in /var/log/messages: Nov 19 14:51:10 ast kernel: SysRq : Show Regs I waited a few minutes, then tried the 't': [EMAIL PROTECTED] init.d]# echo t > /proc/sysrq-trigger which seemed to hang, so I killed it about thirty seconds later, and now my /var/log/messages has 20,000 extra lines :):) I grepped for the PID and found this: Nov 19 14:52:40 ast kernel: modprobe R running 2988 17744 1 31140 28078 (NOTLB) The next line started with 'sshd', so I guess there was no trace with this? > BTW: what kernel? What ditsribution? Keep in mind it has been running almost 1000 days ;) [EMAIL PROTECTED] init.d]# uname -a Linux ast.jbtelenet.com 2.6.9-22.0.2.ELsmp #1 SMP Thu Jan 5 17:13:01 EST 2006 i686 i686 i386 GNU/Linux I believe it is Redhat 9. Its a colo... Thanks for the interesting debug pointers! j _______________________________________________ -- Bandwidth and Colocation Provided by http://www.api-digital.com -- asterisk-users mailing list To UNSUBSCRIBE or update options visit: http://lists.digium.com/mailman/listinfo/asterisk-users
