CODEWITHSUNDEEP
pythonscopeglobal-variablesvariable-declaration
deepbrook
deepbrook

Reputation: 2656

Accessing a variable from outer scope via global keyword in Python

As per this Question's accepted answer, I understand I cannot create pure global variables. Okay, cool.

However, he then goes on and says:

[..]all you can do is make a variable in a particular scope. (If you make a variable inside the Python interpreter, and then import other modules, your variable is in the outermost scope and thus global within your Python session.[..]

Okay, so we can't assign globals in the original sense, but it appears to be possible to access variables in the outermost scope from within a package, via the global keyword, correct?

I am then, apparently, missing something crucial in my efforts to access a variable passed to my python program via commandline arguments.

My program has the usual __main__.py, which handles argument parsing and executes code from my python module backend.

backend has code that relies on input via command line arguments. However, I seem to fail at making these arguments available to backend.

My package's layout is:

mypackage
- __main__.py
- backend/
    -__init__.py
    -direction.py

Here's __main__.py:

import argparse

# Setup ArgParser
parser = argparse.ArgumentParser(description="Fancy Description of program.")
parser.add_argument('--tar', nargs=1, dest='target',
                help='Specify output folder.',
                default=['/path/to/output'])

target_dir = args.target[0]

# Import backend now that our variable is set.
from backend.direction import direction

And my backend/direction.py:

global target_dir
print(target_dir)

Running this raises a NameError: 'target_dir' is not defined. So where's the poop ? Am I assuming an impossibility here, or am I simply messing up on declaration?

Upvotes: 7

Views: 21835

Answers (2)

dodell
dodell

Reputation: 490

I am not clear what your problem actually is. Is it not possible to do the following?

import backend

if __name__ == "__main__":
    # parse arguments
    backend.argreceive(arguments)

As Daniel Roseman suggested, if there are many backend functions requiring access to these variables then you should consider using a class, and class properties, to store the variables.

class argreceive():
    def __init__(self,args):
        self.args = args

Upvotes: 0

Brendan Abel
Brendan Abel

Reputation: 37509

global only works within the module it's used in. You also don't use global to declare a global variable at the global scope (ie. the module scope). You would use global within the scope of a function to indicate that you want to use a variable in the global scope and not in the local scope.

In python 3, there is also the nonlocal keyword that allows you to indicate you want to modify a variable in an outer scope that isn't the global/module scope.

global

A = 1

def global_func():
    global A
    A = 2


def func():
    A = 3


print(A)
# 1

global_func()
print(A)
# 2

func()
print(A)
# 2

nonlocal

def outside():
    a = 1

    def inside_nonlocal():
        nonlocal a
        a = 2

    def inside():
        a = 3

    print(a)
    # 1

    inside_nonlocal()
    print(a)
    # 2

    inside()
    print(a)
    # 2

Upvotes: 20

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