[IPython-dev] [FWD] IPythonShellEmbed() disturbs regular (embedded) Python traceback

[Fernando Perez]

Forwarded to the list, please note that you need to be subscribed to post (we 
get way too much spam otherwise).

I'll reply on-list, sorry for the delay, this slipped in a pile of other email.

Cheers,

f


# ORIGINAL MESSAGE:

Subject:
IPythonShellEmbed() disturbs regular (embedded) Python traceback
From:
"Remy X.O. Martin" <rmvsxop at gmail.com>
Date:
Sat, 30 Sep 2006 15:13:24 +0200
To:
ipython-dev at scipy.org

Hello,

I am embedding Python in one of my C apps, and use IPythonShellEmbed
to support an interactive Python shell, using

      if( !called ){
           PyRun_SimpleString( "from IPython.Shell import IPythonShellEmbed" );
           PyRun_SimpleString( "xgraph.ipshell=IPythonShellEmbed()" );
           PyRun_SimpleString( "xgraph.ipshell.set_banner('Entering
xgraph.ipshell IPython shell')" );
           PyRun_SimpleString( "xgraph.ipshell.set_exit_msg('Leaving
xgraph.ipshell IPython shell')" );
            // Not clear if this is needed or actually does anything:
           PyRun_SimpleString( "xgraph.ipshell.restore_system_completer()" );
           called= 1;
      }


If I call this upon initialising the embedded interpreter, (some)
syntax errors in user code no longer generate useful error messages:

def RedrawNow:
      return None

---------------------------------------------------------------------------
None                                      Traceback (most recent call last)


None: None


If I do *not* initialise the interactive shell, I get the usual message

def RedrawNow:
      return None

   File "<string>", line 1
     def RedrawNow:
                  ^
SyntaxError: invalid syntax


This is of course quite annoying: execution of the Python code stops
at the offending statement, but there is no way to find out what that
statement was.

To be honest, I don't see why IPython would become active outside an
eventual call to xgraph.ipshell() as defined above...?!

Regards,
R.M.