[lldb-dev] Debug events in synchronous mode?

[Pavel Labath via lldb-dev]

Hi,

I am not sure what are the "official" rules, but the general idea is
that you need not concern yourself too much with events when you are
in synchronous mode. In synchronous mode, you can be sure that by the
time target.Launch() returns, the process will be stopped (or dead, or
something else, you can check process.GetState() to find that out). No
need to play around with listeners...

cheers,
pl

On 4 February 2016 at 06:41, Jeffrey Tan via lldb-dev
<lldb-dev at lists.llvm.org> wrote:
> Hi,
>
> I found that if I am using synchronous mode, some times there are no debug
> events generated during launch. For example, for the code below,
> LLDBListenerThread will receive no debug events.
>
> Is this expected? What is the rule of debug events in synchronous mode?
>
> def main():
>     debugger = lldb.SBDebugger.Create()
>     debugger.SetAsync(False)
>     target = debugger.CreateTargetWithFileAndArch(executable_path,
> lldb.LLDB_ARCH_DEFAULT)
>     target.BreakpointCreateByName('main')
>
>     listener = lldb.SBListener('Event Listener')
>     error = lldb.SBError()
>     process = target.Launch (listener,
>                              None,      # argv
>                              None,      # envp
>                              None,      # stdin_path
>                              None,      # stdout_path
>                              None,      # stderr_path
>                              None,      # working directory
>                              0,         # launch flags
>                              False,     # Stop at entry
>                              error)     # error
>     print 'Launch result: %s' % str(error)
>     event_thread = LLDBListenerThread(debugger)
>     event_thread.start()
>
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