How to assign new value to a variable using multiprocessing in python

Question

def foo(var, new_value):
    var = new_value

x = 1
multiprocessing.Process(target=foo, args=(x,2))
p.start()
p.join()
print(x)

I would expect the output of the print(x) is 2, but in fact I got 1. How could I assign new values to an existing variable?

Your foo() function doesn't change x, because x is fully evaluated before being passed to the function. var receives the value of x in var. You then change var to a different value. var is a local variable, so it goes away when foo() ends. — kindall
– kindall, Commented Aug 6, 2018 at 19:06
Adding to @kindall: The important part here is that this has nothing to do with multiprocessing; this wouldn't work if you just did foo(x, 2) with no multiprocessing involved at all. — ShadowRanger
– ShadowRanger, Commented Aug 6, 2018 at 19:28

CAB · Accepted Answer · 2018-08-06 19:20:36Z

1

You need to use multiprocessing shared values;

import multiprocessing
import ctypes

def foo(var, new_value):
    var.value = new_value

x = multiprocessing.Value(ctypes.c_int, 1)
print(x.value)
p = multiprocessing.Process(target=foo, args=(x,2))
p.start()
p.join()
print(x.value)

answered Aug 6, 2018 at 19:20

CAB

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1 Comment

ShadowRanger Over a year ago

Note that this is changing two things: 1. It uses shared memory so multiprocessing can transfer data from child to a location where the parent can see it. 2. It changes a rebinding of var in foo to an attribute assignment; rebinding can never change the caller's version of an argument, but modifying an attribute will change the caller's value. #2 isn't specific to multiprocessing; it's a general rule of Python.

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