How to get path of a .py script

[Skip Montanaro]

    Manuel> if __name__ == '__main__':
    Manuel>     p = sys.argv[0]
    Manuel> else:
    Manuel>     p = __file__

This test won't be necessary in 2.3.  Consider:

    % cat > f.py
    import sys
    print sys.argv[0]
    print __file__
    % python f.py
    f.py
    f.py

This is with current CVS.  Just an FYI.

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