Nonlocal

We say that a variable defined in a frame is local to that frame. A variable is nonlocal to a frame if it is defined in the environment that the frame belongs to but not the frame itself, i.e. in its parent or ancestor frame.

So far, we know that we can access variables in parent frames:

def make_adder(x):
    """ Returns a one-argument function that returns the result of
    adding x and its argument. """
    def adder(y):
        return x + y
    return adder

Here, when we call make_adder, we create a function adder that is able to look up the name x in make_adder's frame and use its value.

However, we haven't been able to modify variable in parent frames. Consider the following function:

def make_withdraw(balance):
    """Returns a function which can withdraw 
    some amount from balance
    """

    >>> withdraw = make_withdraw(50)
    >>> withdraw(25)
    25
    >>> withdraw(25)
    0
    """
    def withdraw():
        if amount > balance:
            return "Insufficient funds"
        balance = balance - amount
        return balance
    return withdraw

The inner function withdraw attempts to update the variable balance in its parent frame. Running this function's doctests, we find that it causes the following error:

UnboundLocalError: local variable 'balance' referenced before assignment

Why does this happen? When we execute an assignment statement, remember that we are either creating a new binding in our current frame or we are updating an old one in the current frame. For example, the line balance = ... in withdraw, is creating the local variable balance inside withdraw's frame. This assignment statement tells Python to expect a variable called balance inside withdraw's frame, so Python will not look in parent frames for this variable. However, notice that we tried to compute balance - amount before the local variable was created! That's why we get the UnboundLocalError.

To avoid this problem, we introduce the nonlocal keyword. It allows us to update a variable in a parent frame!

Some important things to keep in mind when using nonlocal

• nonlocal cannot be used with global variables (names defined in the global frame).
• If no nonlocal variable is found with the given name, a SyntaxError is raised.
• A name that is already local to a frame cannot be declared as nonlocal.

Consider this improved example:

def make_withdraw(balance):
    """Returns a function which can withdraw 
    some amount from balance
    """

    >>> withdraw = make_withdraw(50)
    >>> withdraw(25)
    25
    >>> withdraw(25)
    0
    """
    def withdraw(amount):
        nonlocal balance
        if amount > balance:
            return "Insufficient funds"
        balance = balance - amount
        return balance
    return withdraw

The line nonlocal balance tells Python that balance will not be local to this frame, so it will look for it in parent frames. Now we can update balance without running into problems.

Mutability

We say that an object is mutable if its state can change as code is executed. The process of changing an object's state is called mutation. Examples of mutable objects include lists and dictionaries. Examples of objects that are not mutable include tuples and functions.

We have seen how to use the == operator to check if two expressions evaluate to equal values. We now introduce a new comparison operator, is, that checks whether two expressions evaluate to the same values.

Wait, what's the difference? For primitive values, there is none:

>>> 2 + 2 == 3 + 1
True
>>> 2 + 2 is 3 + 1
True

This is because all primitives have the same identity under the hood. However, with non-primitive values, such as lists, each object has its own identity. That means you can construct two objects that may look exactly the same but have different identities.

>>> lst1 = [1, 2, 3, 4]
>>> lst2 = [1, 2, 3, 4]
>>> lst1 == lst2
True
>>> lst1 is lst2
False

Here, although the lists referred to by lst1 and lst2 have equal contents, they are not the same object. In other words, they are the same in terms of equality, but not in terms of identity.

This is important in our discussion of mutability because when we mutate an object, we simply change its state, not its identity.

>>> lst1 = [1, 2, 3, 4]
>>> lst2 = lst1
>>> lst1.append(5)
>>> lst2
[1, 2, 3, 4, 5]
>>> lst1 is lst2
True