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Home/ Questions/Q 6163789
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T21:53:00+00:00 2026-05-23T21:53:00+00:00

So, I’ve got a multithreaded python program, which is currently suffering from deadlock. I

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So, I’ve got a multithreaded python program, which is currently suffering from deadlock. I was going to log lock acquiring by subclassing threading.Lock objects:

import traceback
class DebugLock(threading.Lock):
    def acquire(self):
        print >>sys.stderr, "acquired", self
        #traceback.print_tb
        threading.Lock.acquire(self)  
    def release(self):
        print >>sys.stderr, "released", self
        #traceback.print_tb
        threading.Lock.release(self)  

When I try to run the program, I get the following error:

    class DebugLock(threading.Lock):
TypeError: Error when calling the metaclass bases
    cannot create 'builtin_function_or_method' instances  

So, my question is twofold:

  1. Is it possible to subclass Lock objects to do what I’m doing?

  2. If not, what is the best way to debug deadlock in python?

Note: I’m not writing any Python extension. There’s a similar question: How to debug deadlock with python?
However, it deals with compiling C++ code and using GDB, which I can’t do since my code is pure python.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T21:53:01+00:00Added an answer on May 23, 2026 at 9:53 pm

    You could just use the "has a lock" versus "is a lock" approach, like so:

    import threading, traceback, sys
    class DebugLock(object):
        def __init__(self):
            self._lock = threading.Lock()
        def acquire(self):
            print("acquired", self)
            #traceback.print_tb
            self._lock.acquire()
        def release(self):
            print("released", self)
            #traceback.print_tb
            self._lock.release()
        def __enter__(self):
            self.acquire()
        def __exit__(self, type, value, traceback):
            self.release()
    

    where I’ve thrown in the appropriate context guards since you likely want to use the with syntax with your locks (who wouldn’t?).

    Usage shown below:

        >>> lock = DebugLock()
        >>> with lock:
        ...     print("I'm atomic!")
        ... 
        acquired <__main__.DebugLock object at 0x7f8590e50190>
        I'm atomic!
        released <__main__.DebugLock object at 0x7f8590e50190>
        >>>
    
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