weird namespace phenomenon #35220

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1. nobody commented on Sep 21, 2001

   

   nobodymannequin

   on Sep 21, 2001

   Mannequin

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   # == module f.py
   y = 10
   
   def g():
   
     if y:
       y = 100
   
     return y

   # == module h.py

   import f
   print f.g()

   # == plop these two modules into the same directory
   and 'cd' to that directory.

   When you run 'python h.py' at the command line, line 6
   of f.py (if y:)
   creates a problem due to f being supposedly undefined.
   The error is
   registered as NameError or UnboundLocalError depending
   on what version
   of the python core you're running. Python 2.1.x says
   y is being
   referenced before assignment, yet if you change the
   _next_ line to read
   "z = 100" rather than "y = 100", the whole thing runs
   fine, printing 10
   (the value of f.y) at the end as expected.

   This is at the very least an extreme logic quirk in
   how python namespaces
   work, and at worst a bug badly in need of correction
   (since vsn 1.5.2 or
   before).
2. anonymous

   closed this as completedon Sep 21, 2001
3. anonymous

   added

   invalid

   on Sep 21, 2001
4. anonymous

   assigned

   tim-one

   on Sep 21, 2001
5. anonymous

   added

   interpreter-core(Objects, Python, Grammar, and Parser dirs)(Objects, Python, Grammar, and Parser dirs)

   on Sep 21, 2001
6. anonymous

   closed this as completedon Sep 21, 2001
7. anonymous

   added

   invalid

   on Sep 21, 2001
8. anonymous

   assigned

   tim-one

   on Sep 21, 2001
9. anonymous

   added

   interpreter-core(Objects, Python, Grammar, and Parser dirs)(Objects, Python, Grammar, and Parser dirs)

   on Sep 21, 2001
10. tim-one commented on Sep 21, 2001

    

    tim-one

    on Sep 21, 2001

    Member

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    Logged In: YES
    user_id=31435

    Whether a given name is local or global in a given scope is
    a property determined at compile-time, not at runtime. See
    the Reference Manual for details. In brief, if a name is
    bound ("assigned to") *anywhere* in the body of a function,
    that name is local throughout the entire function. You
    assign to y in the body of g(), therefore y is a local
    variable everywhere in g(), and you reference y's value in g
    () before giving it a value (hence it's an unbound local at
    the time you first reference it, hence UnboundLocalError in
    recent Pythons). The y in the body of g() has nothing to
    do with the module-level vrbl y. If you intended to
    reference the module-level vrbl y, then you need to put

    global y

    as the first statement in g().

    But it's impossible for me to guess what you intended --
    and it's also impossible for Python.

    If you're familiar with C, it's exactly the same as

    int y = 10;
    
    int g() {int y; if (y) y = 100; return y;}

    except that in C you'd get random crap out of that function
    (due to referencing an uninitialized auto) instead of a
    reliable exception. The difference is that Python doesn't
    have explict declarations for local variables, but the fact
    that you assigned to y within the body of g makes y a local
    just as much as y is function-local in the C example above.
11. ezio-melotti

    transferred this issue fromon Apr 9, 2022