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Home/ Questions/Q 8872045
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T18:07:47+00:00 2026-06-14T18:07:47+00:00

Possible Duplicate: Least Astonishment in Python: The Mutable Default Argument I have this code

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Possible Duplicate:
“Least Astonishment” in Python: The Mutable Default Argument

I have this code

class Test(object):
  def __init__(self, var1=[]):
    self._var1 = var1

t1 = Test()
t2 = Test()

t1._var1.append([1])

print t2._var1

and I get “[[1]]” as the result. So clearly t1._var1 and t2._var1 are addressing the same list. If I put

t3 = Test()
print t3._var1

then I get “[[1]]” as well. So var1=[] seems to permanently bind var1 to the some list. I tried copying the list,

def __init__(self, var1=copy([])):

but got the same result, so the expression for the named argument appears to be evaluated prior to init being called, and it just gave var1 a copy of the empty list which was then shared amongst the instances.

So how do I use [] as a default value for a named argument?

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  1. Editorial Team
    Editorial Team
    2026-06-14T18:07:49+00:00Added an answer on June 14, 2026 at 6:07 pm

    You can’t use [] directly if you want each object to have an empty list. I tend to use a work around:

    def __init__(self, var1=None):
        if var1 is None:
            var1 = []
        ....
    

    Naturally this won’t work if var1 can be None, you would need to use a different object.

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