# The Mutable Default Argument Mess in Python

Meet Kevin, 🙋‍♂️ Kevin is learning Python. One day he was given a problem to solve as follows:

> Design a function that appends '#' to a list provided as an argument and then prints it. If no argument is provided then the function should use an empty list as a default.

Kevin quickly comes up with the following solution:

```python
def append(l = []):
    l.append('#')
    print(l)
```

Looks good, time for testing the solution:

```python
append([1, 2, 3])
# OUTPUT: [1, 2, 3, '#'] | OK

append()
# OUTPUT: ['#']	| OK

append()
# OUTPUT: ['#', '#']  | Strange!!
```

- The first call to `append` worked fine. It added **#** to the list `[1, 2, 3]` and then printed it.

- The second also worked as expected. This time no list is provided as an argument, so it used the default empty list and appended a **#** to it.

- Now, the third call resulted in something unexpected.

  When we once again called the `append` without argument, it printed `['#', '#']` instead of `['#']` as we got in the above call.



# Why did this happened?

The reason this happened is that Python defines its default argument *only once when the function is first defined.*

This is because python is parsed line by line, when the parser encounters `def` it sets the default argument to a value and that value is used in every future call.

This behavior of Python becomes of special concern when the default argument is a **mutable**.

As the value of **, immutables** can not be changed, if you update the argument variable inside the function it will create a new object and start pointing to that object instead of changing the original default object.

But in the case of a mutable default argument, the object created at the time of parsing the function is updated instead of creating a different object for that function call.



## Solution

The solution to this problem is to use an immutable default argument instead of a mutable. The preferred choice is `None` (though you can choose any immutable value).

```python
def append(l = None):
    if l is None:
        l = []
    l.append('#')
    print(l)
```

Let's test this solution -

```python
append([1, 2, 3])
# OUTPUT: [1, 2, 3, '#']	| OK

append()
# OUTPUT: ['#'] 	| OK

append()
# OUTPUT: ['#']		| Works fine!
```

Great! this solution worked as expected.

But why? Let's take a look inside..



## Why the solution works?

Watch this video to find out what happened in the **wrong version** of the code -

%[https://youtu.be/NAEQd0HFxyQ]

As you can see, in this case, the original `l` was modified instead of creating a new `l`  for each function call as a `list` is a mutable value.



Now, see the **corrected version** of the code -

%[https://youtu.be/sQWhLwnAh04]

Here as `None` is an immutable value therefore it cannot be changed and a new list object is created for each function call.

> Note: The default argument is a property of the **function object** therefore, initially it is the same for all function calls.

### Thanks For Reading 😊

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Feedbacks are welcome in the comments.

---



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