CODEWITHSUNDEEP
pythongarbage-collection
Casebash
Casebash

Reputation: 118782

Manual garbage collection in Python

Is there any way to manually remove an object which the garbage collection refuses to get rid of even when I call gc.collect()? Working in Python 3.0

Upvotes: 20

Views: 37946

Answers (4)

Alex Martelli
Alex Martelli

Reputation: 881575

Per the docs, gc.get_referrers(thatobject) will tell you why the object is still alive (do it right after a gc.collect() to make sure the undesired "liveness" is gonna be persistent). After that, it's somehow of a black art;-). You'll often find that some of the referrers are lists (so WHY is that list referring to thatobject? you can .remove it in an emergency mode, but making the normal code sound is better...), and, even more often, dicts (many of whose may be __dict__s of some class instance or other -- often not trivial to find out which one... again, brute-force removal is sometimes an expedient emergency solution, but never a sustainable long-range one!-).

Upvotes: 29

Tzury Bar Yochay
Tzury Bar Yochay

Reputation: 9004

del Or None are your only friends

>>> a = "Hello"
>>> a = None
Or
>>> del a

Upvotes: 4

paxdiablo
paxdiablo

Reputation: 881303

If the GC is refusing to destroy it, it's because you have a reference to it somewhere. Get rid of the reference and it will (eventually) go. For example:

myRef = None

Keep in mind that GC may not necessarily destroy your object unless it needs to.

If your object is holding resources not under the management of Python (e.g., some trickery with C code called from Python), the object should provide a resource release call so you can do it when you want rather than when Python decides.

Upvotes: 4

Bostone
Bostone

Reputation: 37126

It depends on what your Python is running on. Here's good article that explains the details

Quoting:

In current releases of CPython, each new assignment to x inside the loop will release the previously allocated resource. Using GC, this is not guaranteed. If you want to write code that will work with any Python implementation, you should explicitly close the resource; this will work regardless of GC:

for name in big_list:
    x = Resource()
    do something with x
    x.close()

Upvotes: 1

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