Creating distinct objects in a function

Question

Take a look to the following code snippet:

class MyObj(object):
    name = ""

    def __init__(self, name):
        self.name = name

v = [ {} ] * 2

def f(index):
    v[index]['surface'] = MyObj('test')
    v[index]['num'] = 3


if __name__ == '__main__':
    f(0)
    f(1)

    v[0]['num'] = 4
    print v[1]['num']

What I expected to get as output of the last line is a 3; however it prints out 4. So it should mean that the new object is created always at the same reference address.

How can I avoid this behaviour? (i.e. how can I make the above code prints 4?)

Recommended stackoverflow.com/questions/240178/… A near dupe — Bhargav Rao
– Bhargav Rao, Commented May 19, 2015 at 21:32

Padraic Cunningham · Accepted Answer · 2015-05-19 21:46:33Z

5

You need to create two dicts:

v = [ {},{} ]

Or use a loop:

v = [ {} for _ in range(2)]

You are creating a two references to the same object.

In [2]: a = [{}] * 2

In [3]: id(a[0])
Out[3]: 140209195751176

In [4]: id(a[1])
Out[4]: 140209195751176

In [5]: a[0] is a[1]
Out[5]: True

In [6]: a = [{} for _ in range(2)]  

In [7]: id(a[1])
Out[7]: 140209198435720    

In [8]: id(a[0])
Out[8]: 140209213918728 

In [9]: a[0] is a[1]
Out[9]: False

edited May 19, 2015 at 21:46

answered May 19, 2015 at 21:31

Padraic Cunningham

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6 Comments

Bhargav Rao Over a year ago

@AmiTavory Note that Whenever you see [ {} ] * 2 or [ [] * 0] It is THE culprit

Ami Tavory Over a year ago

@Bhargav Rao Unfortunately, my brain's lexer is slower than some other peoples' bytecode compiler. :-)

Bhargav Rao Over a year ago

@AmiTavory You better compile this answer into a pyc file and store it in your persistent storage. :P

Padraic Cunningham Over a year ago

@AmiTavory, I learned everything I know from the master Bhargav Rao ;)

Padraic Cunningham Over a year ago

@BhargavRao, I dunno,I have been told I have an honset face ;)

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