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Shared memory in multiprocessing

I have three large lists. First contains bitarrays (module bitarray 0.8.0) and the other two contain arrays of integers.

l1=[bitarray 1, bitarray 2, ... ,bitarray n]
l2=[array 1, array 2, ... , array n]
l3=[array 1, array 2, ... , array n]

These data structures take quite a bit of RAM (~16GB total).

If i start 12 sub-processes using:

multiprocessing.Process(target=someFunction, args=(l1,l2,l3))

Does this mean that l1, l2 and l3 will be copied for each sub-process or will the sub-processes share these lists? Or to be more direct, will I use 16GB or 192GB of RAM?

someFunction will read some values from these lists and then performs some calculations based on the values read. The results will be returned to the parent-process. The lists l1, l2 and l3 will not be modified by someFunction.

Therefore i would assume that the sub-processes do not need and would not copy these huge lists but would instead just share them with the parent. Meaning that the program would take 16GB of RAM (regardless of how many sub-processes i start) due to the copy-on-write approach under linux? Am i correct or am i missing something that would cause the lists to be copied?

EDIT: I am still confused, after reading a bit more on the subject. On the one hand Linux uses copy-on-write, which should mean that no data is copied. On the other hand, accessing the object will change its ref-count (i am still unsure why and what does that mean). Even so, will the entire object be copied?

For example if i define someFunction as follows:

def someFunction(list1, list2, list3):
    i=random.randint(0,99999)
    print list1[i], list2[i], list3[i]

Would using this function mean that l1, l2 and l3 will be copied entirely for each sub-process?

Is there a way to check for this?

EDIT2 After reading a bit more and monitoring total memory usage of the system while sub-processes are running, it seems that entire objects are indeed copied for each sub-process. And it seems to be because reference counting.

The reference counting for l1, l2 and l3 is actually unneeded in my program. This is because l1, l2 and l3 will be kept in memory (unchanged) until the parent-process exits. There is no need to free the memory used by these lists until then. In fact i know for sure that the reference count will remain above 0 (for these lists and every object in these lists) until the program exits.

So now the question becomes, how can i make sure that the objects will not be copied to each sub-process? Can i perhaps disable reference counting for these lists and each object in these lists?

EDIT3 Just an additional note. Sub-processes do not need to modify l1, l2 and l3 or any objects in these lists. The sub-processes only need to be able to reference some of these objects without causing the memory to be copied for each sub-process.

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Answer

Generally speaking, there are two ways to share the same data:

  • Multithreading
  • Shared memory

Python’s multithreading is not suitable for CPU-bound tasks (because of the GIL), so the usual solution in that case is to go on multiprocessing. However, with this solution you need to explicitly share the data, using multiprocessing.Value and multiprocessing.Array.

Note that usually sharing data between processes may not be the best choice, because of all the synchronization issues; an approach involving actors exchanging messages is usually seen as a better choice. See also Python documentation:

As mentioned above, when doing concurrent programming it is usually best to avoid using shared state as far as possible. This is particularly true when using multiple processes.

However, if you really do need to use some shared data then multiprocessing provides a couple of ways of doing so.

In your case, you need to wrap l1, l2 and l3 in some way understandable by multiprocessing (e.g. by using a multiprocessing.Array), and then pass them as parameters.
Note also that, as you said you do not need write access, then you should pass lock=False while creating the objects, or all access will be still serialized.

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