UnboundLocalError in nested function #53295

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1. AndreasHofmeister commented on Jun 21, 2010

   

   AndreasHofmeistermannequin

   on Jun 21, 2010

   MannequinAuthor

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   Description:
   An unexpected UnboundLocalError is produced when assigning a value to a variable inside a nested function. The first assignment to the variable is in the enclosing function.

   Example:
   def x():
   a = False
   def y():
   print a
   a = True
   return y

   Calling x()() produces an UnboundLocalError on the 'print a' line.
   If the 'a = True' line is removed, no error occurs.

   Tested with:

   • 2.5.1
   • 2.6.5

   Keywords:
   Nested function, UnboundLocalError, variable assignment

   Thank you for your attention
2. AndreasHofmeister

   added

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

   type-bugAn unexpected behavior, bug, or errorAn unexpected behavior, bug, or error

   on Jun 21, 2010
3. mdickinson commented on Jun 21, 2010

   

   mdickinson

   on Jun 21, 2010

   Member

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   This isn't a bug; it's by design.

   Because there's an assignment to 'a' in the function 'y', 'a' is considered local to that function. (It doesn't matter where the assignment happens within the function; the presence of an assignment anywhere is enough to make 'a' local for the entirety of 'y'.)

   This is described in the reference manual at:

   http://docs.python.org/reference/executionmodel.html#naming-and-binding

   See the paragraph beginning:

   "If a name binding operation occurs anywhere within a code block, "
4. mdickinson

   closed this as completedon Jun 21, 2010
5. mdickinson

   added

   invalid

   on Jun 21, 2010
6. AndreasHofmeister commented on Jun 22, 2010

   

   AndreasHofmeistermannequin

   on Jun 22, 2010

   MannequinAuthor

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   Thank you for your assistance. I apologize for not examining the reference manual closely.
   Is there any way to produce the desired behavior? I currently work around the local name binding like this:

   def x():
        a = [False]
        def y():
            print a[0]
            a[0] = True
        return y

   However, using a list here seems awkward.
   Thank you for your attention.
7. mdickinson commented on Jun 22, 2010

   

   mdickinson

   on Jun 22, 2010

   Member

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   Is there any way to produce the desired behavior?

   Not directly, in Python 2.x. (But there's the 'nonlocal' keyword in 3.x.) There are various workarounds, but what's best depends on what you're doing. The python-list mailing list is probably a better place to get answers.
8. ezio-melotti

   transferred this issue fromon Apr 10, 2022