UnboundLocalError with nested function scopes [duplicate]

This question already has answers here: Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope? (9 answers) Closed 4 months ago.

I have code like this (simplified):

def outer():
    ctr = 0

    def inner():
        ctr += 1

    inner()

But ctr causes an error:

Traceback (most recent call last):
  File "foo.py", line 9, in <module>
    outer()
  File "foo.py", line 7, in outer
    inner()
  File "foo.py", line 5, in inner
    ctr += 1
UnboundLocalError: local variable 'ctr' referenced before assignment

How can I fix this? I thought nested scopes would开发者_C百科 have allowed me to do this. I've tried with 'global', but it still doesn't work.

---

If you're using Python 3, you can use the nonlocal statement to enable rebinding of a nonlocal name:

def outer():
    ctr = 0

    def inner():
        nonlocal ctr
        ctr += 1

    inner()

If you're using Python 2, which doesn't have nonlocal, you need to perform your incrementing without barename rebinding (by keeping the counter as an item or attribute of some barename, not as a barename itself). For example:

...
ctr = [0]

def inner():
    ctr[0] += 1
...

and of course use ctr[0] wherever you're using bare ctr now elsewhere.

---

The Explanation

Whenever a value is assigned to a variable inside a function, python considers that variable a local variable of that function. (It doesn't even matter if the assignment is executed or not - as long as an assignment exists in a function, the variable being assigned to will be considered a local variable of that function.) Since the statement ctr += 1 includes an assignment to ctr, python thinks that ctr is local to the inner function. Consequently, it never even tries to look at the value of the ctr variable that's been defined in outer. What python sees is essentially this:

def inner():
    ctr = ctr + 1

And I think we can all agree that this code would cause an error, since ctr is being accessed before it has been defined.

(See also the docs or this question for more details about how python decides the scope of a variable.)

The Solution (in python 3)

Python 3 has introduced the nonlocal statement, which works much like the global statement, but lets us access variables of the surrounding function (rather than global variables). Simply add nonlocal ctr at the top of the innerfunction and the problem will go away:

def outer():
    ctr = 0

    def inner():
        nonlocal ctr
        ctr += 1

    inner()

The Workaround (in python 2)

Since the nonlocal statement doesn't exist in python 2, we have to be crafty. There are two easy workarounds:

• Removing all assignments to ctr

  Since python only considers ctr a local variable because there's an assignment to that variable, the problem will go away if we remove all assignments to the name ctr. But how can we change the value of the variable without assigning to it? Easy: We wrap the variable in a mutable object, like a list. Then we can modify that list without ever assigning a value to the name ctr:

  def outer():
      ctr = [0]
  
      def inner():
          ctr[0] += 1
  
      inner()
  

• Passing ctr as an argument to inner

  def outer():
      ctr = 0
  
      def inner(ctr):
          ctr += 1
          return ctr
  
      ctr = inner(ctr)
  

---

How about declaring ctr outside of outer (i.e. in the global scope), or any other class/function? This will make the variable accessible and writable.