Where is the default parameter in Python function

Question

I think many people have seen the python’s function which receives default parameters. For example:

def foo(a=[]):
    a.append(3)
    return a

If we call this function using foo(), the output will append integer ‘3’ each time after the call.

When this function is defined, a function object named ‘foo’ is defined in the current environment, and also the default parameter values are evaluated at this time. Every time when the function is called without a parameter, the evaluated parameter value will be changed according to the code.

My question is, where is this evaluated parameter exist?
Is it in the function object or it is in the method object when calling the function?
Since everything in python is a object, there must be some place to hold the name->value binding of ‘a’–>evaluated parameter. Am I over-thinking this problem?

Asked By: ming.kernel

||

Source

Answer 1

It’s attached to the function object, see foo.func_defaults:

>>> foo()
>>> foo.func_defaults
([3],)
>>> foo()
>>> foo.func_defaults
([3, 3],)

In case if you want to get the mapping of a onto [], you can access foo.func_code:

defaults = foo.func_defaults
# the args are locals in the beginning:
args = foo.func_code.co_varnames[:foo.func_code.co_argcount] 
def_args = args[-len(defaults):]  # the args with defaults are in the end
print dict(zip(def_args, defaults)) # {'a': []}

(But, apparently, the yak’s version is better.)

Answered By: bereal

Answer 2

It’s in the function object, in the func_defaults:

def f(a=[]): a.append(3)

print f.func_defaults # ([],)

f()

print f.func_defaults # ([3],)

Answered By: rplnt

Answer 3

It is stored in the func_defaults attribute of the function object.

>>> foo.func_defaults
([],)
>>> foo()
([3],)

Answered By: rkhayrov

Answer 4

As others already said, the default values are stored in the function object.

For example, in CPython you can do this:

>>> def f(a=[]):
...     pass
...
>>> f.func_defaults
([],)
>>> f.func_code.co_varnames
('a',)
>>>

However, co_varnames may contain more than names of args so it needs further processing and these attributes might not even be there in other Python implementations. Therefore you should use the inspect module instead which takes care of all implementation details for you:

>>> import inspect
>>> spec = inspect.getargspec(f)
>>> spec
ArgSpec(args=['a'], varargs=None, keywords=None, defaults=([],))
>>>

The ArgSpec is a named tuple so you can access all values as attributes:

>>> spec.args
['a']
>>> spec.defaults
([],)
>>>

As the documentation says, the defaults tuple always corresponds to the n last arguments from args. This gives you your mapping.

To create a dict you could do this:

>>> dict(zip(spec.args[-len(spec.defaults):], spec.defaults))
{'a': []}
>>>

Answered By: yak

Answer 5

I have found an interesting situation: in python 2.5.2 version, try the function ‘foo()’

>>> foo()
[1]
>>> foo()
[1]
>>> foo()
[1]

Because the objects of the function called are different:

>>> id(foo())
4336826757314657360
>>> id(foo())
4336826757314657008
>>> id(foo())
4336826757314683160

In 2.7.2 version:

>>> foo()
[1]
>>> foo()
[1, 1]
>>> foo()
[1, 1, 1]

In this case, the object is the same each time calling the function:

>>> id(foo())
29250192
>>> id(foo())
29250192
>>> id(foo())
29250192

Is it a problem of different versions?

Answered By: zfz

Answer 6

In Python 3, funcdefaults is __defaults__:

>>> def foo(a=[]):
...     a.append(3)
...     return a
... 
>>> foo.__defaults__
([],)

Answered By: Loren

Where is the default parameter in Python function

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