bpython -i & namespaces

Question

I can't seem to find this answer anywhere.

Given the trivial example:

# myclass.py
class MyClass:
    def __init__(self):
        print 'test'

def main():
    my_class_instance = MyClass()

if __name__ == '__main__':
    main()

some_var = 'i exist!  ... but I think I'm in global namespace?'

If I run bpython -i myclass.py, I execute the program & drop into the bpython environment. Whatever namespace I'm in - my_class_instance does not exist. However, some_var does exist - and so does the main function itself.

Is there anyway that I can pull any objects that exist in that main function into the namespace I'm in when I drop into that interactive prompt? Or is there something else I should do?

Answer 1

Using interactive bpython over python 2.7, __name__ is equal to __console__. So your function main() may be never called. A hack would be to write:

# myclass.py
class MyClass:
    def __init__(self):
        print 'test'

def main():
    global my_class_instance
    my_class_instance = MyClass()

if __name__ in ('__main__', '__console__'):
    main()

Answer 2

my_class_instance is in main's namespace, so you can't see it outside of main. Use a global instead:

my_class_instance = None

def main():
    global my_class_instance

    my_class_instance = MyClass()

Answer 3

Another trick I sometimes use when I want something back from running a main function is to return what I need at the end of main.

So, for example if I need an instance and some other variable from the main function in the top level one could do:

def main():
    myclass = MyClass()
    a = 4
    return (my class, a)

if __name__ == '__main__':
    ret = main()

If you now call your script with bpython -i scriptname you will have the variable 'ret' in the global namespace and ret[0] has your class instance, ret[1] has the number 4.