I don't understand this python __del__ behaviour

python del
Brett Stottlemyer picture Brett Stottlemyer · May 24, 2011 · Viewed 50.1k times · Source

Can someone explain why the following code behaves the way it does:

import types

class Dummy():
    def __init__(self, name):
        self.name = name
    def __del__(self):
        print "delete",self.name

d1 = Dummy("d1")
del d1
d1 = None
print "after d1"

d2 = Dummy("d2")
def func(self):
    print "func called"
d2.func = types.MethodType(func, d2)
d2.func()
del d2
d2 = None
print "after d2"

d3 = Dummy("d3")
def func(self):
    print "func called"
d3.func = types.MethodType(func, d3)
d3.func()
d3.func = None
del d3
d3 = None
print "after d3"

The output (note that the destructor for d2 is never called) is this (python 2.7)

delete d1
after d1
func called
after d2
func called
delete d3
after d3

Is there a way to "fix" the code so the destructor is called without deleting the method added? I mean, the best place to put the d2.func = None would be in the destructor!

Thanks

[edit] Based on the first few answers, I'd like to clarify that I'm not asking about the merits (or lack thereof) of using __del__. I tried to create the shortest function that would demonstrate what I consider to be non-intuitive behavior. I'm assuming a circular reference has been created, but I'm not sure why. If possible, I'd like to know how to avoid the circular reference....

Answer

Nick Bastin picture Nick Bastin · May 24, 2011

You cannot assume that __del__ will ever be called - it is not a place to hope that resources are automagically deallocated. If you want to make sure that a (non-memory) resource is released, you should make a release() or similar method and then call that explicitly (or use it in a context manager as pointed out by Thanatos in comments below).

At the very least you should read the __del__ documentation very closely, and then you should probably not try to use __del__. (Also refer to the gc.garbage documentation for other bad things about __del__)

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