Default parameter value for objects

Question

Python provides a way to set a default value for function parameters. An example is:

def f(x=3):
    print(x)

This is for a primitive type, lets try with objects:

def f(x=list()):
    print(id(x))     
f()

44289920

f()

44289920

Same object! I was surprised of this being used to the C/C++ way. Done with that, I now understand the default value is not build at invoking time but at definition time.

So I came to a solution:

def f(x=list()):
    if len(x) == 0:
        x = list()
    print(id(x))

Solved! But at what price: In my opinion this doesn't seem to be a very clean solution. This solution rely in the use of len(x) == 0 as a way to identify the default value which is Ok for my function but not for others so the solution can be generalized as:

def f(x=None):
    if x is None:
        x = list()

This can be shortened to:

def f(x=None):
    x = x or list()  # a bit shorter version

My question is, is there any shorter or better way to solve this problem? Will it ever be?

x=list() is executed exactly once when the def is executed. So the default value for x of all calls to f refer to the same object. — user1907906
– user1907906, Commented Jun 27, 2014 at 9:00
The question is: "is there any shorter or better way to solve this problem? Will it ever be?". The answer is: No, because (see above). — user1907906
– user1907906, Commented Jun 27, 2014 at 9:01
Related: Why the “mutable default argument fix” syntax is so ugly, asks python newbie — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Jun 27, 2014 at 9:05
Only improvement you can do here is to use x is None instead of ==. — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Jun 27, 2014 at 9:08

denisvm · Accepted Answer · 2014-07-18 05:54:11Z

2

I still prefer the is None approach, but here is a new option to think about: If is defined only the type, you create a new instance of it.

def f(x=list):
    if isinstance(x, type): x = x()
    print(id(x))

answered Jul 18, 2014 at 5:54

denisvm

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