destructors, exceptions and the lock guard paradigm

Ron Samuel Klatchko moocat94122 at yahoo.com
Wed Jun 18 04:25:29 CEST 2003


I am trying to use the lock guard paradigm in Python.  My first
attempt was
this:

class guard:
    def __init__(self, lock):
        self.acquired = 0
        self.lock = lock
        self.lock.acquire()
        self.acquired = 1

    def __del__(self):
        print "releasing lock"
        self.release()

    def release(self):
        if self.acquired:
            self.lock.release()
            self.acquired = 1

Intended to be used like this:

class someclass:
    def __init__(self):
        self.lock = threading.Lock()

    def somefunc(self):
        g = guard(self.lock)
        # do some work while the lock is held

The idea was that when somefunc exited, the reference count to g would
drop to 0, the guard would be destructed which would release the lock.
 That way, I do not have to manually release the lock for every exit
path.

While it works for normal returns from the function, the lock is not
released when an exception is thrown.  From my research it looks to be
due to how the last traceback is stored and continues to hold a
reference to the guard object.

Is there a way to get this to work under Python?  I realize I can use
try/finally block and put the call to Lock.release() there, I don't
like that as much as the guard method.

samuel




More information about the Python-list mailing list