Let's look at a common Python wat and try to figure out wat's actually happening!
We'll define a function, foo, which takes one argument, l, which has the default value of an empty list.
>>> def foo(l=[]):
... l.append('cat')
... return l
What happens when we call foo multiple times?
>>> foo()
['cat']
>>> foo()
['cat', 'cat']
Whoa! So mutating l actually mutates it for all future calls to the function. Weird.
This means that the [] object is only created once, and each time we call foo without an argument, l is referring to that same object. This may lead you to form a hypothesis: l=[] is kind of like a name-binding statement that executes only once, when the function is defined.
But, if that hypothesis is true, then how should we expect the following function to behave?
>>> def bar(l=[]):
... print locals()
... l = ['cat']
... return l
...
>>> bar()
# ?
>>> bar()
# ?
Well, if l=[] is like a name-binding statement that executes only once when the function is defined, then I would expect something like this sequence of events to happen, when we define bar and then call it twice:
bar is defined
- the name
l is bound to the object []
bar is called the first time:
locals() should return {l : []}- then we reassign
l to ['cat'] within the scope of bar
bar should return ['cat']
bar is called again:
l=[] is not executed (based on our assumption)locals() should either return {} or {l : ['cat']}, depending on if the assignment of l = ['cat'] persists after the function is called the first timebar should return ['cat']
What actually happens?
>>> bar()
{'l': []}
['cat']
>>> bar()
{'l': []}
['cat']
Hrm. This behavior reasonably leads us to believe that the assignment l=[] happens each time we call the function bar. But in foo, l=[] can't be a statement that executes each time the function is called, or else we'd create a new [] each time.
If we assume l=[] executes like a name-binding statement, then it must execute either (1) only once when the function is defined, or (2) each time the function is called. In foo, it only executes once, but in bar, it executes every time we call bar. That just can't be. So our assumption that l=[] executes like a name-binding statement leads to a contradiction, and thus must be wrong!
Guess what, nerds! We kind of just did a proof by contradiction!
So then what really happens when we define default values for arguments? Let's see if we can figure out where the default values are stored.
My usual go-to for questions like this is Python internals whiz and Hacker School Facilitator Allison Kaptur, but you can also find the answer with a bit of googling.
So, without further ado, what actually happens when we define a default argument in Python 2.x is that the value of the argument gets stored inside the function's func_defaults method. (In 3.x, the values are stored in the __defaults__ method.)
Let's look back at the foo function:
>>> def foo(l=[]):
... l.append('cat')
... return l
In Python 2.x, we can access foo's func_defaults like so:
>>> foo.func_defaults
([],)
>>> foo()
['cat']
>>> foo.func_defaults
(['cat'],)
Aha! So the actual object that is being stored as the default for foo is being modified when we call foo! For fun, let's see if we can mutate the default value from outside of the function:
>>> foo.func_defaults[0].append('dragon')
>>> foo.func_defaults
(['cat', 'dragon'],)
>>> foo()
['cat', 'dragon', 'cat']
Eep! That was fun. So what's in the func_defaults of bar? Recall:
>>> def bar(l=[]):
... print locals()
... l = ['cat']
... return l
And then:
>>> bar.func_defaults
([],)
>>> bar()
{'l': []}
['cat']
>>> bar.func_defaults
([],)
Okay! So since bar reassigns l to ['cat'], it doesn't modify the object stored in func_defaults.
So what have we learned?
It appears as if the following happens when we define and call bar:
bar is defined
- the object
[] is created and stored in the func_defaults tuple
bar is called the first time:
- since we didn't pass in a value for
l as an argument, Python looks in the func_defaults for the value to bind to the name l, and grabs the [] object that we created when we defined bar
locals() returns {l : []}- we reassign
l to ['cat'] within the scope of bar. Since this is a reassignment, this doesn't modify the [] object contained in func_defaults. Instead, l is just bound to a different object in memory.
['cat'] is returned
bar is called again:
- since we didn't modify the
[] object the first time we called bar, the same series of events happens as in step 2!
I should probably also mention something more useful: a common way of setting a default value to an empty list (and having it actually work as expected) is to do the following:
>>> def baz(l=None):
... if not l:
... l = []
... l.append('cat')
... return l
When we call baz multiple times, its behavior is more expected:
>>> baz()
['cat']
>>> baz()
['cat']
Cool! Wat conquered. We should collect badges for all the wats we've battled.
Thanks to Mary Rose Cook and Allison Kaptur, who valiantly battled this wat with me.
tags:
python
python internals
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