Don t understand why UnboundLocalError occurs closure

Question

What am I doing wrong here   counter   0  def increment      counter    1  increment     The above code throws an UnboundLocalError

User · Answer

Python has lexical scoping by default  which means that although an enclosed scope can access values in its enclosing scope  it cannot modify them  unless they re declared global with the global keyword    A closure binds values in the enclosing environment to names in the local environment  The local environment can then use the bound value  and even reassign that name to something else  but it can t modify the binding in the enclosing environment   In your case you are trying to treat counter as a local variable rather than a bound value  Note that this code  which binds the value of x assigned in the enclosing environment  works fine    gt  gt  gt  x   1   gt  gt  gt  def f     gt  gt  gt   return x   gt  gt  gt  f   1

User · Answer

Python is not purely lexically scoped   See this  Using global variables in a function  and this  https   www saltycrane com blog 2008 01 python-variable-scope-notes

User · Answer

The reason of why your code throws an UnboundLocalError is already well explained in other answers   But it seems to me that you re trying to build something that works like itertools count      So why don t you try it out  and see if it suits your case    gt  gt  gt  from itertools import count  gt  gt  gt  counter   count 0   gt  gt  gt  counter count 0   gt  gt  gt  next counter  0  gt  gt  gt  counter count 1   gt  gt  gt  next counter  1  gt  gt  gt  counter count 2

User · Answer

Python doesn t have variable declarations  so it has to figure out the scope of variables itself   It does so by a simple rule   If there is an assignment to a variable inside a function  that variable is considered local  1   Thus  the line  counter    1   implicitly makes counter local to increment     Trying to execute this line  though  will try to read the value of the local variable counter before it is assigned  resulting in an UnboundLocalError  2   If counter is a global variable  the global keyword will help   If increment   is a local function and counter a local variable  you can use nonlocal in Python 3 x

User · Answer

To modify a global variable inside a function  you must use the global keyword   When you try to do this without the line  global counter   inside of the definition of increment  a local variable named counter is created so as to keep you from mucking up the counter variable that the whole program may depend on   Note that you only need to use global when you are modifying the variable  you could read counter from within increment without the need for the global statement

User · Answer

You need to use the global statement so that you are modifying the global variable counter  instead of a local variable   counter   0  def increment      global counter   counter    1  increment     If the enclosing scope that counter is defined in is not the global scope  on Python 3 x you could use the nonlocal statement   In the same situation on Python 2 x you would have no way to reassign to the nonlocal name counter  so you would need to make counter mutable and modify it   counter    0   def increment      counter 0     1  increment   print counter 0     prints  1

User · Answer

try this  counter   0  def increment      global counter   counter    1  increment

User · Answer

To answer the question in your subject line   yes  there are closures in Python  except they only apply inside a function  and also  in Python 2 x  they are read-only  you can t re-bind the name to a different object  though if the object is mutable  you can modify its contents   In Python 3 x  you can use the nonlocal keyword to modify a closure variable   def incrementer        counter   0     def increment            nonlocal counter         counter    1         return counter     return increment  increment   incrementer    increment       1 increment       2     The question origially asked about closures in Python

[python] Don't understand why UnboundLocalError occurs (closure)

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