Python Functions Defaults Explained

I have been on the several Python interviews again and have passed an Upwork Python test. And I have noticed that the interviewers like using the task as the following one.

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def f(x, l=[]):
    for i in range(x):
        l.append(i * i)
    return l
>>> f(2)
>>> f(3, [0, 1, 2])
>>> f(3)

Question: what is the output of those lines?

The output of the first two lines is pretty obvious, but the result of the third line f(3) wasn’t so intuitive for me.

So let’s investigate what is going on after the initialization of the f function. I use IPython to run this code.

>>> f
<function __main__.f(x, l=[])>
>>> f.__defaults__
([],)

The empty list that we see from the f.__defaults__ result is basically the l variable in the function code.

>>> f(2)
[0, 1]

Nothing special.

>>> f
<function __main__.f(x, l=[0, 1])>
>>> f.__defaults__
([0, 1],)

But! Now, we can see that variable l has a [0, 1] value because of the mutability of the Python list object and passing function arguments as a reference.

>>> f(3, [0, 1, 2])
[0, 1, 2, 0, 1, 4]
>>> f
<function __main__.f(x, l=[0, 1])>