Open riyadparvez opened 1 week ago
arg_name=True
requires the source code access but ipython definitely does something weird when it enters the debugger. If you throw out arg_name=True
this should work fine.
Thank you for the workaround. That being said I do think an exception shouldn't be thrown since this is primarily a debugging tool. Is it possible to just catch the exception and leave the source line empty?
Yes, TypeError
should be caught together with OSError
if the source is not available. I will fix it later. Thanks for the issue.
On first hit of the breaskpoint if I continue and on the second hit of the
op(i)
throws exception.The traceback: