global variables in imported modules

Discussion in 'Python' started by vsoler, May 16, 2010.

  1. vsoler

    vsoler Guest

    Taken from www.python.org, FAQ 2.3 How do I share global variables
    across modules?

    config.py:

    x = 0 # Default value of the 'x' configuration setting


    mod.py:

    import config
    config.x = 1


    main.py:

    import config # try removing it
    import mod
    print config.x

    The example, such as shown in the website, works perfectly well.
    However, I don't fully understand why I have to import config in
    main.py, since it has already been imported by mod.py.

    As the website explains, there is only one module namespace for each
    module, and mod.py has aleady created the config namespace by
    importing it. Why should I import it again in main.py if that
    namespace already exists?

    If I remove -> import config # try removing it in main.py,
    the application does not run

    What am I missing?
     
    vsoler, May 16, 2010
    #1
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  2. What you are missing is that the interpreter has to look *inside* a
    namespace in order to actually find the object associated with a
    name. As you found out, there is a namespace per module. So
    main.py's namespace is where the code in main.py will search for
    variables. If 'mod' imports config, then the 'mod' module's namespace
    is updated with 'config' -> the config module. But the act of 'mod'
    importing 'config' will not alter the namespace of 'main' at all. So
    if you want to access variable 'x' inside 'config' from main you can
    either import config directly into main and access it as config.x, or
    you can import config into mod and import mod into main and access it
    as mod.config.x.

    Regards,
    Pat
     
    Patrick Maupin, May 16, 2010
    #2
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  3. vsoler

    vsoler Guest

    Thank you Pat, it's very clear.

    However, can I be 100% sure that,no matter how I access variable
    'x' (with config.x or mod.config.x) it is always the same 'x'. I mean
    that either reference of 'x' points to the same id(memory position)?

    Thank you
     
    vsoler, May 16, 2010
    #3
  4. vsoler

    James Mills Guest

    Yes it does unless you re-assign it.

    --James
     
    James Mills, May 16, 2010
    #4
  5. vsoler

    vsoler Guest

    Understood, thank you very much

    Vicente Soler
     
    vsoler, May 16, 2010
    #5

  6. To expand a bit on what James is saying:

    If, for example, inside your main module, you got tired of typing
    "mod.config.x" everywhere you were using it, and decided that you
    could make a local reference to the same variable:

    x = mod.config.x

    Now, whenever you use just plain x inside the main module, you are
    also referencing the exact same object, *until* some other function
    decides to do:

    mod.config.x = y

    At this point in time, the 'x' inside main references the object that
    mod.config.x originally referenced, but mod.config.x now references a
    different object.

    Unlike C, for example, where the assignment operator physically places
    an item into a specific memory location (either fixed globally or
    within a stack frame), the assignment operator in python simply stores
    a key/value pair into a namespace dictionary. So whenever you
    retrieve a value from the dictionary using that key, you will get the
    value that was last associated with that key.

    So, 'mod.config.x' will first retrieve the object associated with the
    key 'mod' from the main module's namespace dictionary, then will
    retrieve the object associated with the key 'config' from that
    module's namespace dictionary, then will retrieve the object
    associated with the key 'x' from that module's namespace dictionary.
    Unless you later modify any of those key/value pairs, subsequent
    retrieval will always result in the same final value.

    Regards,
    Pat
     
    Patrick Maupin, May 16, 2010
    #6
  7. vsoler

    vsoler Guest

    Really interesting, it helps a lot.

    Thank you
     
    vsoler, May 17, 2010
    #7
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