How to access (and edit) variables from a callback function?

Question

I use Boto to access Amazon S3. And for file uploading I can assign a callback function. The problem is that I cannot access the needed variables from that callback function until I make them global. In another hand, if I make them global, they are global for all other Celery tasks, too (until I restart Celery), as the file uploading is executed from a Celery task.

Here is a function that uploads a JSON file with information about video conversion progress.

def upload_json():
    global current_frame
    global path_to_progress_file
    global bucket
    json_file = Key(bucket)
    json_file.key = path_to_progress_file
    json_file.set_contents_from_string('{"progress": "%s"}' % current_frame,
    cb=json_upload_callback, num_cb=2, policy="public-read")

And here are 2 callback functions for uploading frames generated by ffmpeg during the video conversion and a JSON file with the progress information.

# Callback functions that are called by get_contents_to_filename.
# The first argument is representing the number of bytes that have
# been successfully transmitted from S3 and the second is representing
# the total number of bytes that need to be transmitted.
def frame_upload_callback(transmitted, to_transmit):
    if transmitted == to_transmit:
        upload_json()
def json_upload_callback(transmitted, to_transmit):
    global uploading_frame
    if transmitted == to_transmit:
        print "Frame uploading finished"
        uploading_frame = False

Theoretically, I could pass the uploading_frame variable to the upload_json function, but it wouldn’t get to json_upload_callback as it’s executed by Boto.

In fact, I could write something like this.

In [1]: def make_function(message):
   ...:     def function():
   ...:         print message
   ...:     return function
   ...: 

In [2]: hello_function = make_function("hello")

In [3]: hello_function
Out[3]: <function function at 0x19f4c08>

In [4]: hello_function()
hello

Which, however, doesn’t let you edit the value from the function, just lets you read the value.

def myfunc():
    stuff = 17
    def lfun(arg):
        print "got arg", arg, "and stuff is", stuff
    return lfun

my_function = myfunc()
my_function("hello")

This works.

def myfunc():
    stuff = 17
    def lfun(arg):
        print "got arg", arg, "and stuff is", stuff
        stuff += 1
    return lfun

my_function = myfunc()
my_function("hello")

And this gives an UnboundLocalError: local variable 'stuff' referenced before assignment.

Thanks.

Answer 1

In Python 2.x closures are read-only. You can however use a closure over a mutable value... i.e.

def myfunc():
    stuff = [17] # <<---- this is a mutable object
    def lfun(arg):
        print "got arg", arg, "and stuff[0] is", stuff[0]
        stuff[0] += 1
    return lfun

my_function = myfunc()
my_function("hello")
my_function("hello")

If you are instead using Python 3.x the keyword nonlocal can be used to specify that a variable used in read/write in a closure is not a local but should be captured from the enclosing scope:

def myfunc():
    stuff = 17
    def lfun(arg):
        nonlocal stuff
        print "got arg", arg, "and stuff is", stuff
        stuff += 1
    return lfun

my_function = myfunc()
my_function("hello")
my_function("hello")