python - class mutability

Question

I got the code below from an exam and I don't understand why the first time when you make f2 = f1, doing f1.set() changes f2 but after that when you set f1 = Foo("Nine", "Ten") doesn't change f2 at all. If anyone knows why please explain it to me. Thank you so much!

Code:

class Foo():
    def __init__(self, x=1, y=2, z=3):
        self.nums = [x, y, z]

    def __str__(self):
        return str(self.nums)

    def set(self, x):
        self.nums = x

f1 = Foo()
f2 = Foo("One", "Two")

f2 = f1
f1.set(["Four", "Five", "Six"])
print f1
print f2

f1 = Foo("Nine", "Ten")
print f1
print f2

f1.set(["Eleven", "Twelve"])
print f1
print f2

Outcome:

['Four', 'Five', 'Six']
['Four', 'Five', 'Six']
['Nine', 'Ten', 3]
['Four', 'Five', 'Six']
['Eleven', 'Twelve']
['Four', 'Five', 'Six']

Does this old answer of mine help? [Python list doesn't reflect variable change, new to python](http://stackoverflow.com/questions/12080552/12080644#12080644) — Martijn Pieters, Dec 11 '12 at 23:58
Try running this code through the visualizer at [Python Tutor](http://www.pythontutor.com). — Marius, Dec 12 '12 at 00:14

score 5 · Answer 1 · answered Dec 12 '12 at 00:00

f2 = f1

After this statement both f1 and f2 are references to the same instance of Foo. Therefore, a change in one will effect the other.

f1 = Foo("Nine", "Ten")

After this, f1 is assigned to a new Foo instance so f1 and f2 are no longer connected in any way - so a change in one will not effect the other.

python - class mutability

1 Answers1