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?

yak · Accepted Answer · 2012-04-12 10:15:24Z

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': []}
>>>

bereal · Accepted Answer · 2012-04-12 10:14:54Z

5

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.)

edited Apr 12, 2012 at 10:14

answered Apr 12, 2012 at 9:36

bereal

34.7k8 gold badges65 silver badges111 bronze badges

1 Comment

MattH Over a year ago

Yes, but where is the mapping of a to the [] as the OP has asked?

rplnt · Accepted Answer · 2012-04-12 09:37:39Z

4

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 Apr 12, 2012 at 9:37

rplnt

2,41917 silver badges14 bronze badges

1 Comment

MattH Over a year ago

Yes, but where is the mapping of a to the [] as the OP has asked?

rkhayrov · Accepted Answer · 2012-04-12 09:37:44Z

2

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

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

answered Apr 12, 2012 at 9:37

rkhayrov

10.3k2 gold badges38 silver badges42 bronze badges

Comments

zfz · Accepted Answer · 2012-04-12 14:31:01Z

1

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 Apr 12, 2012 at 14:31

zfz

1,6472 gold badges24 silver badges46 bronze badges

1 Comment

yak Over a year ago

Shouldn't happen. Make sure you're using exactly the same functions on both versions.

Loren · Accepted Answer · 2023-02-01 17:37:07Z

0

In Python 3, funcdefaults is __defaults__:

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

answered Feb 1, 2023 at 17:37

Loren

13.6k4 gold badges44 silver badges50 bronze badges

Collectives™ on Stack Overflow

Where is the default parameter in Python function

6 Answers 6

Comments

1 Comment

1 Comment

Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

1 Comment

1 Comment

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related