Re: [PATCH 1/1] ptrace: make sure do_wait() won't hang afterPTRACE_ATTACH

[Jan Kratochvil]

On Sun, 20 Feb 2011 18:06:30 +0100, Denys Vlasenko wrote:
> On Sunday 20 February 2011 10:40, Jan Kratochvil wrote:
> > Sure by default GDB does not do anything special, it will respawn (using
> > PTRACE_CONT(SIGSTOP)) any SIGSTOP it sees due to the default setting of:
> > 	(gdb) handle SIGSTOP
> > 	Signal        Stop  Print Pass to program Description
> > 	SIGSTOP       Yes Yes Yes   Stopped (signal)
> > 
> > Therefore there happens the double SIGSTOP reporting as discussed before:
> > 	(gdb) run
> > 	Starting program: /bin/sleep 1h
> > 	# external kill -STOP <inferior pid>
> > 	Program received signal SIGSTOP, Stopped (signal).
> > 	# State:	t (tracing stop)
> > 	(gdb) continue 
> > 	Continuing.
> > 	Program received signal SIGSTOP, Stopped (signal).
> > 	# State:	t (tracing stop)
> > 	(gdb) continue 
> > 	Continuing.
> > 	# State:	S (sleeping)
> > 
> > Your proposal is I expect:
> > 	(gdb) run
> > 	Starting program: /bin/sleep 1h
> > 	# external kill -STOP <inferior pid>
> > 	Program received signal SIGSTOP, Stopped (signal).
> > 	# State:	t (tracing stop)
> > 	(gdb) continue 
> > 	Continuing.
> > 	# State:	T (stopped)
> 
> Not exactly. Even after we fix kernel so that it properly preserves
> group-stop across ptrace-stops, gdb will still see TWO 
> waitpid:SIGSTOP events, not one.
> 
> First one says "the tracee has received SIGSTOP", and after PTRACE_CONT(SIGSTOP),
> second one says "the tracee has stopped because of SIGSTOP".
> Currently, neither strace nor gdb understands that second one
> is different from first.

I thought the kernel change being discussed should be the second SIGSTOP will
receive the process's original parent, so that CTRL-Z on a debugged process
from shell works the normal way, as without a debugger.


Thanks,
Jan
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