weakref and thread safety (in python 2.1)
Duncan Booth
duncan at NOSPAMrcp.co.uk
Tue Jul 22 08:53:34 EDT 2003
"Ames Andreas (MPA/DF)" <Andreas.Ames at tenovis.com> wrote in
news:mailman.1058872336.14143.python-list at python.org:
> I've noted that, when thread2 incidentally blocks on the queue
> (because it's full), while thread1 deletes the queue, the weak
> reference isn't deleted and thread2 keeps blocking forever.
That's because when the only way to use a weak reference is to convert it
into a strong reference. So when you call:
myweakref.put(item)
you create a strong reference that exists until the put method returns (and
in fact some more references are created such as the self parameter inside
the put method).
>
> I came up with the following destruction scheme for the queue (within
> thread1; q is the 'strong' ref to the queue), which *seems* to solve
> my problem (as of some preliminary tests):
>
> qw = weakref.proxy(q) # a second weak reference (this time within
> # thread1)
> del q
> try:
> qw.get_nowait() # if thread2 blocks on the (weakrefed) queue,
> # qw still exists here; the get_nowait() call
> # wakes thread2 up. I *hope* that its
> # weakreference will be destroyed when it can
> # block again in its next call to put()
> # (leading to a weakref.ReferenceError)
> except:
> pass
>
> Is this thread-safe? I don't know enough about the implementation of
> the weakref module to decide if it is guaranteed that thread2's weak
> reference to the queue will be destroied *before* thread2 can call
> 'put()' (or rather before thread2 can block) for the next time.
No, this isn't thread safe. You have again got a strong reference while
calling the method, so the only place the weak reference could be destroyed
is outside the method call, and by that time thread2 could have blocked on
the queue again.
However, if you repeatedly try to pull data out of the queue, you should
eventually get somewhere.
qw = weakref.ref(q) # a second weak reference (this time within
# thread1)
del q
while qw:
qw().get_nowait()
There is still a problem though as your code is free-running: if the queue
gets empty before it is released you may use a lot of CPU before it
eventually gets freed. Also you are depending on the memory behaviour of C
Python, if Zope ever gets ported to Jython you will probably find that weak
references don't go away until the garbage collector kicks in.
A much better way to do this would be to make the termination explicit.
e.g. (untested code)
class MyQueue(Queue):
def __init__(self, maxsize=0):
Queue.__init__(self, maxsize)
self.terminated = False
thread2, with queue a normal reference to a MyQueue instance:
while 1:
item = produce()
queue.put(item)
if queue.terminated:
break # Stop processing queue
thread1 can then terminate the queue with:
queue.terminated = True
queue.get_nowait() # Ensure any blocked put completes.
No weak references needed.
--
Duncan Booth duncan at rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?
More information about the Python-list
mailing list