Ok, this still explains why the instruction leads to a SIGSEGV. What I
don´t understand is why gdb finds a valid pointer to pthread_mutex_t:
----8<-----
Core was generated by `corosync'.
Program terminated with signal 11, Segmentation fault.
#0 0x00007f7ff68078e9 in pthread_mutex_lock () from
/usr/lib/libpthread.so.1
(gdb) bt
#0 0x00007f7ff68078e9 in pthread_mutex_lock () from
/usr/lib/libpthread.so.1
#1 0x00007f7ff7002e14 in ipc_thread_active (conn=0x7f7ff7391000) at
coroipcs.c:466
#2 pthread_ipc_consumer (conn=0x7f7ff7391000) at coroipcs.c:675
#3 0x00007f7ff6809d75 in ?? () from /usr/lib/libpthread.so.1
#4 0x00007f7ff60759f0 in ___lwp_park50 () from /usr/lib/libc.so.12
#5 0x00007f7feb400000 in ?? ()
#6 0x00007f7ff7ec0540 in ?? ()
#7 0x0000000111110001 in ?? ()
#8 0x0000000033330003 in ?? ()
#9 0x0000000000000000 in ?? ()
(gdb) frame 1
#1 0x00007f7ff7002e14 in ipc_thread_active (conn=0x7f7ff7391000) at
coroipcs.c:466
466 pthread_mutex_lock (&conn_info->mutex);
(gdb) print &conn_info->mutex
$1 = (pthread_mutex_t *) 0x7f7ff7391050
(gdb) p *$
$2 = {ptm_magic = 858980355, ptm_errorcheck = 0 '\000', ptm_pad1 =
"\000\000", ptm_interlock = 0 '\000', ptm_pad2 = "\000\000", ptm_owner =
0x0,
ptm_waiters = 0x0, ptm_recursed = 0, ptm_spare2 = 0x0}
--------------------
The disassemly of the pthread_mutex_lock function (till the instruction
that causes the crash) is this:
--------8<--------
(gdb) x/2i 0x00007f7ff68078e0
0x7f7ff68078e0 <pthread_mutex_lock>: mov %fs:0x0,%rax
=> 0x7f7ff68078e9 <pthread_mutex_lock+9>: mov 0x10(%rax),%rdx
---------------
I´m not sure what these instructions are supposed to do - i guess they
should work on the parameter.
Any idea how I can track this down?
2012/12/5 Manuel Bouyer <[email protected]>
> On Wed, Dec 05, 2012 at 02:21:51PM +0100, Stephan wrote:
> > Hi Manuel,
> >
> > thanks, that makes sense. The failing instruction is this one
> >
> > => 0x7f7ff68078e9 <pthread_mutex_lock+9>: mov 0x10(%rax),%rdx
> >
> > with rax containing a pointer to 0x7f7ffffffffe
> >
> > I wonder how many bytes will be written to memory - my imagination is 8
> > bytes as this is a 64bit processor.
>
> No, with mov this will be 4 bytes (an int). For a movl would do a 8 bytes
> write.
>
> --
> Manuel Bouyer <[email protected]>
> NetBSD: 26 ans d'experience feront toujours la difference
> --
>
