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Re: [PATCH 0/9 V3] Use reinsert breakpoint for vCont;s


On Mon, Nov 14, 2016 at 02:10:32PM -0500, Antoine Tremblay wrote:
> 
> > I do something slightly differently in V3.  In my
> > "V2 Use reinsert breakpoint for vCont;s", I install reinsert breakpoints
> > for needed lwps in two places, linux_resume and proceed_all_lwps, which
> > isn't ideal.
> >
> > After the chat with Pedro, we don't need to stop all threads when inserting
> > reinsert breakpoint, so we can move the breakpoint installation further
> > down to linux_resume_one_thread and proceed_one_lwp.
> 
> I'm following up on random SIGILL/SIGSEGV when using software single stepping/
> range stepping with GDBServer on ARM.
> 
> And I can't see why we don't need to stop all threads when inserting
> reinsert breakpoint.
> 
> Since linux_resume will call:
> 
> find_inferior (&all_threads, linux_resume_one_thread,
>   &leave_all_stopped);
> 
> This will start one thread after the other. So for example if thread 3
> has a single step breakpoint to install this will start thread 1, then
> thread 2 and just modify the program's memory to install reinsert
> breakpoints on thread 3 with thread 1 and 2 running.
> 
> Thus leading to thread 1 or 2 executing invalid memory, thus the SIGILL
> random problems...

Single-step breakpoint is thread specific, so we don't need to stop
other threads when inserting one for a specific thread.  Given the
example above, we insert single-step breakpoint for thread 3 on address
A, if thread 1 goes through address A, but doesn't hit the breakpoint,
IOW, thread 1 still sees the original instruction, that is nothing wrong,
right?  We don't expect thread 1 hits that breakpoint for thread 3 anyway.
If thread 1 hits the breakpoint (IOW, thread 1 sees the breakpoint
instruction), GDBserver just handles that SIGTRAP, and it
has already know that there is a breakpoint on address A.

Thread 1 either sees the original instruction on address A or the
breakpoint instruction.  Unless ptrace read/write 32-bit is not
atomic, IOW, partial ptrace write result is visible to other
threads, I don't see why we get SIGILL here.

Note that we stop all threads when we remove single-step breakpoints
because we want no thread sees single-step breakpoint in memory from
their point of view afterwards.

-- 
Yao (齐尧)


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