Mutable Default Arguments in Python

Objects of built-in types like (int, float, bool, str, tuple, Unicode) are immutable. Objects of built-in types like (list, set, dict) are mutable. A mutable object can change its state or contents and immutable objects cannot.

This is a very simple definition of a Mutable and Immutable object in Python. Now coming to the interesting part which is Mutable Default Object in function definitions.

I have written a very simple function as below -

def foobar(element, data=[]):
    data.append(element)
    return data

In the above function, I set data to a list(Mutable object) as a default argument. Now let’s execute this function -

>>> print(foobar(12))
[12]

The output is as expected, Now execute multiple times -

>>> print(foobar(22))
[12, 22]
>>> print(foobar(32))
[12, 22, 32]

What is going on here? As you can see, the first time, the function returns exactly what we expected. On the second and all subsequent passes the same list is used in each call.

Why is this happening

Python’s default arguments are evaluated once when the function is defined, not each time the function is called. When Python encounters it, the first thing it will do is compile it in order to create a code object for this function. While this compilation step is done, Python evaluates and then stores the default arguments in the function object itself.

So, let’s do some introspection, to clear the confusions, I have taken a few lines of code as below-

• Function Before Execution -

def foo(l=[]):
    l.append(10)

After function executes this definition, You can check the default attribute for this function. by using __defaults__.

About __defaults__ A tuple containing default argument values for those arguments that have defaults, or None if no arguments have a default value.

>>> foo.__defaults__
([],)

The result an empty list as the only entry in __defaults__

• Function After Execution-

Let’s now execute this function and check the defaults.

>>> foo()
>>> foo.__defaults__
([10],)

Astonished? The value inside the object changes, Consecutive calls to the function will now simply append to that embedded list object:

Let’s execute the function multiple times:

>>> foo()
>>> foo()
>>> foo()

>>> foo.__defaults__
([10, 10, 10, 10],)

The reason why this is happening because default argument values are stored in the function object, not in the code object (because they represent values calculated at run-time).

The Solution

Now, the question is how to code foobar in order to get the behavior that we expected.

Fortunately, the solution is straightforward. The mutable objects used as defaults are replaced by None, and then the arguments are tested for None.

def foobar(element, data=None):
    if data is None:
        data = []
    data.append(element)
    return data

>>> foobar(12)
[12]

So, by replacing mutable default arguments with None solved our problem.

Good Use of Mutable Default Argument

However, Mutable Default Argument have some good use-case also as following-

1.Binding local variable to the current value of the outer variable in a callback

2.Cache/Memoization

I hope you like the explanation of Mutable Default Arguments in Python. Still, if any doubt or improvement regarding it, ask in the comment section.

References:

https://docs.python.org/3/reference/datamodel.html#the-standard-type-hierarchy

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