The untimely dimise of a weak-reference
in [Python]
|
Prev: Builtn super() function. How to use it with multiple inheritance? And why should I use it at all?
Next: Basic Information about Python
From: Vincent van Beveren on 30 Jul 2010 08:22 Hi everyone, I was working with weak references in Python, and noticed that it was impossible to create a weak-reference of bound methods. Here is a little python 3.0 program to prove my point: import weakref print("Creating object...") class A(object): def b(self): print("I am still here") a = A() def d(r): print("Aaah! Weakref lost ref") print("Creating weak reference") r = weakref.ref(a.b, d) print("Oh, wait, its already gone!") print("Ref == None, cause of untimely demise: %s" % r()) print("Object is still alive: %s" % a) print("Function is still exists: %s" % a.b) print("See:") a.b() I also tried this in Python 2.5 and 2.6 (with minor modifications to the syntax of course), and it yielded the exact same behavior. Why is this, and is there anything I can do about it? I wish to reference these bound functions, but I do not want to keep them in memory once the object they belong to is no longer referenced. Regards, Vincent van Beveren ___ Ing. V. van Beveren Software Engineer, FOM Rijnhuizen E: V.vanBeveren(a)rijnhuizen.nl
From: Peter Otten on 30 Jul 2010 09:06 Vincent van Beveren wrote: > Hi everyone, > > I was working with weak references in Python, and noticed that it was > impossible to create a weak-reference of bound methods. Here is a little > python 3.0 program to prove my point: > > import weakref > > print("Creating object...") > class A(object): > > def b(self): > print("I am still here") > > a = A() > > def d(r): > print("Aaah! Weakref lost ref") > > print("Creating weak reference") > > r = weakref.ref(a.b, d) The instance doesn't keep a reference of its bound method. Rather the bound method keeps a reference of its instance. Every time you say a.b you get a different bound method. What do you think should keep it alive? > print("Oh, wait, its already gone!") > print("Ref == None, cause of untimely demise: %s" % r()) > print("Object is still alive: %s" % a) > print("Function is still exists: %s" % a.b) > print("See:") > a.b() > > I also tried this in Python 2.5 and 2.6 (with minor modifications to the > syntax of course), and it yielded the exact same behavior. Why is this, > and is there anything I can do about it? I wish to reference these bound > functions, but I do not want to keep them in memory once the object they > belong to is no longer referenced. I fear you have to manage the methods' lifetime explicitly. Peter
From: Vincent van Beveren on 30 Jul 2010 10:06 Hi Peter, I did not know the object did not keep track of its bound methods. What advantage is there in creating a new bound method object each time its referenced? It seems kind of expensive. Regards, Vincent -----Original Message----- From: Peter Otten [mailto:__peter__(a)web.de] Sent: vrijdag 30 juli 2010 15:06 To: python-list(a)python.org Subject: Re: The untimely dimise of a weak-reference Vincent van Beveren wrote: > Hi everyone, > > I was working with weak references in Python, and noticed that it was > impossible to create a weak-reference of bound methods. Here is a little > python 3.0 program to prove my point: > > import weakref > > print("Creating object...") > class A(object): > > def b(self): > print("I am still here") > > a = A() > > def d(r): > print("Aaah! Weakref lost ref") > > print("Creating weak reference") > > r = weakref.ref(a.b, d) The instance doesn't keep a reference of its bound method. Rather the bound method keeps a reference of its instance. Every time you say a.b you get a different bound method. What do you think should keep it alive? > print("Oh, wait, its already gone!") > print("Ref == None, cause of untimely demise: %s" % r()) > print("Object is still alive: %s" % a) > print("Function is still exists: %s" % a.b) > print("See:") > a.b() > > I also tried this in Python 2.5 and 2.6 (with minor modifications to the > syntax of course), and it yielded the exact same behavior. Why is this, > and is there anything I can do about it? I wish to reference these bound > functions, but I do not want to keep them in memory once the object they > belong to is no longer referenced. I fear you have to manage the methods' lifetime explicitly. Peter
From: Peter Otten on 30 Jul 2010 10:18 Vincent van Beveren wrote: > I did not know the object did not keep track of its bound methods. What > advantage is there in creating a new bound method object each time its > referenced? It seems kind of expensive. While I didn't measure it I suppose that it saves a lot of memory. Peter
From: Christian Heimes on 30 Jul 2010 10:44
Am 30.07.2010 16:06, schrieb Vincent van Beveren: > I did not know the object did not keep track of its bound methods. What advantage is there in creating a new bound method object each time its referenced? It seems kind of expensive. Instances of a class have no means of storing the bound method object. The or unbound bound method is a simple and small wrapper that keeps a reference to the class, "self" and the function object. Python keeps a pool of empty method objects in a free list. The creation of a new bound method just takes a few pointer assignments and three INCREFs. Christian |