multiprocessing.sharedctypes and built-in locks

Ahmad Syukri bin Abdollah syockit at gmail.com
Sat Mar 14 13:11:34 CET 2009


I'm trying this on Python 3.0.1
Consider the following code:
"""
import multiprocessing as mp

def jambu(b,i,gl):
    for n in range(100000):
       with gl[i]:
            b[i]+=2
       with gl[3-i]:
            b[3-i]-=1

def main():
    b = mp.RawArray('i',4)
    gl = []
    proc = []
    for i in range(len(b)):
        gl.append(mp.Lock())
        proc.append(mp.Process(target=jambu,args=(b,i,gl)))
    for p in proc:
        p.start()
    for p in proc:
        p.join()
    print(b[:])
    print(sum(b[:]))
main()
"""
(Yes, I'm aware that I didn't pass the lock array as shared variable,
but since they're not reassigned, it should be okay) The above code
should produce an output like this:
  [100000, 100000, 100000, 100000]
  400000
Now, in the documentation for multiprocessing module, it says that
multiprocessing.Array will automatically create a lock to ensure
process-safe synchronization. So I suppose the above code can be
simplified as follows:
"""
import multiprocessing as mp

def jambu(b,i):
    for n in range(100000):
        b[i]+=2
        b[3-i]-=1

def main():
    b = mp.Array('i',4)
    proc = []
    for i in range(len(b)):
        gl.append(mp.Lock())
        proc.append(mp.Process(target=jambu,args=(b,i)))
    for p in proc:
        p.start()
    for p in proc:
        p.join()
    print(b[:])
    print(sum(b[:]))
main()
"""
The output of this second code isn't consistent with the first one,
implying multiprocessing.Array (or even multiprocessing.Value; I've
also tried with this one) isn't as atomic as I understand it should
be. So what is the actual meaning of having "a Lock object which will
be used to synchronize access to the value"? Is it only for getting,
and not for assigning values?

Regards,
Ahmad Syukri



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