Is the variable garbage collected or not?

Question

class X:
    def __init__(self, d={}):
        self.x = []
        d['any'] = self.x

    def append(self, i):
        self.x.append(i)

x = X()
x.append(3)
x.append(13)
x.append(123213)
print(x.x)  # show [3, 13, 123213]

The question is, what happens to d? Is it garbage collected? If it is, how does x still exist?

Why would `d` be garbage collected? You still have `x` and `X` and `X.__init__` in memory. `x` references `X`, and `X` references `__init__` and that function references `d`. — Martijn Pieters, Oct 23 '15 at 14:00
Also the `x` instance references `x.x`, the list, so there is no reason for *that* list to be garbage collected. `x.x` does not reference `d`, that relationship goes the other way. — Martijn Pieters, Oct 23 '15 at 14:01
Last but not least, `d` is a *function default* and stored in the function object. — Martijn Pieters, Oct 23 '15 at 14:02
Regardless of whether or not `d` was done, that would just reduce the reference count of the variables it reverse to, like `x` in this case. Only when a variable's reference count is 0 is it available for garbage collection — Eric Renouf, Oct 23 '15 at 14:02
Don't do this. Never use mutable types as the defaults. Every instance of class `X` will share the same `d` dict. — TheBlackCat, Oct 23 '15 at 14:04
@TheBlackCat Never say never. A mutable type as a default value is perfectly valid, *if* you know that you want to accumulate values across function calls. The trick is knowing when it is appropriate. — chepner, Oct 23 '15 at 14:07

score 3 · Accepted Answer · edited Oct 23 '15 at 14:05

3

By d, you really mean the dict object that serves as the default value for that parameter. There is a "hidden" reference to this dict that prevents it from being garbage collected.

>>> print x.__init__.__defaults__
({'any': [3, 13, 123213]},)

edited Oct 23 '15 at 14:05

hans-t

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answered Oct 23 '15 at 14:02

chepner

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Is the variable garbage collected or not?

1 Answers1