memory allocation about python

Question

I run the following python code.

class A():
    def __init__(self,l=[]):
        self.k = l
    a = A()
    b = A()
    a.k.append(5)
    print id(a.k) == id(b.k)
    print b.k

The result is:

True
[5]

I don't know why a.k and b.k have the same memory address.

Removed totally superfluous k = 1 and reformatted it to look nicer but mostly so it would compile :-) And, please, for the love of whatever god(s) you believe in, consider moving to Python3 - it's so much better. — paxdiablo
– paxdiablo, Commented Jan 22, 2016 at 5:21

paxdiablo · Accepted Answer · 2016-01-22 05:28:41Z

2

Because that empty list you're assigning to l is created once when the function is defined, not each time the function is called.

Hence l for both classes will point to the same list and therefore changing one will change the other (or, more correctly, changing one will change the only one).

If you want to ensure each object of the class gets its own list, simply use the slicing method:

self.k = l[:]

edited Jan 22, 2016 at 5:28

answered Jan 22, 2016 at 5:14

paxdiablo

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