UnboundLocalError in Python
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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.
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'
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 original question's title asked about closures in Python.
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