Why does using `arg=None` fix Python's mutable default argument issue?
It looks like a_list would still be initialized only once
"initialization" is not something that happens to variables in Python, because variables in Python are just names. "initialization" only happens to objects, and it's done via the class' __init__
method.
When you write a = 0
, that is an assignment. That is saying "a
shall refer to the object that is described by the expression 0
". It is not initialization; a
can name anything else of any type at any later time, and that happens as a result of assigning something else to a
. Assignment is just assignment. The first one is not special.
When you write def good_append(new_item, a_list=None)
, that is not "initializing" a_list
. It is setting up an internal reference to an object, the result of evaluating None
, so that when good_append
is called without a second parameter, that object is automatically assigned to a_list
.
meaning a_list would only get reset to [] on the first invocation
No, a_list
gets set to []
any time that a_list
is None
to begin with. That is, when either None
is passed explicitly, or the argument is omitted.
The problem with []
occurs because the expression []
is only evaluated once in this context. When the function is compiled, []
is evaluated, a specific list object is created - that happens to be empty to start - and that object is used as the default.
How does
a_list
get wiped clean every timegood_append
runs?
It doesn't. It doesn't need to be.
You know how the problem is described as being with "mutable default arguments"?
None
is not mutable.
The problem occurs when you modify the object that the parameter has as a default.
a_list = []
does not modify whatever object a_list
previously referred to. It cannot; arbitrary objects cannot magically transform in-place into empty lists. a_list = []
means "a_list
shall stop referring to what it previously referred to, and start referring to []
". The previously-referred-to object is unchanged.
When the function is compiled, and one of the arguments has a default value, that value - an object - gets baked into the function (which is also, itself, an object!). When you write code that mutates an object, the object mutates. If the object being referred to happens to be the object baked into the function, it still mutates.
But you cannot mutate None
. It is immutable.
You can mutate []
. It is a list, and lists are mutable. Appending an item to a list mutates the list.
The problem only exists if the default value is mutable, which None
is not. What gets stored along with the function object is the default value. When the function is called, the function's context is initialized with the default value.
a_list = []
just assigns a new object to the name a_list
in the context of the current function call. It does not modify None
in any way.
The default value of a_list
(or any other default value, for that matter) is stored in the function's interiors once it has been initialized and thus can be modified in any way:
>>> def f(x=[]): return x
...
>>> f.func_defaults
([],)
>>> f.func_defaults[0] is f()
True
resp. for Python 3:
>>> def f(x=[]): return x
...
>>> f.__defaults__
([],)
>>> f.__defaults__[0] is f()
True
So the value in func_defaults
is the same which is as well known inside function (and returned in my example in order to access it from outside.
IOW, what happens when calling f()
is an implicit x = f.func_defaults[0]
. If that object is modified subsequently, you'll keep that modification.
In contrast, an assignment inside the function gets always a new []
. Any modification will last until the last reference to that []
has gone; on the next function call, a new []
is created.
IOW again, it is not true that []
gets the same object on every execution, but it is (in the case of default argument) only executed once and then preserved.