Python nested function scope

Question

I have a python hash that contains a mapping of function name to function. I want to modify each hash entry to call the associated function BUT then to also call a final custom function. It's to act somewhat like an exit hook.

def original():
    print "original work"

becomes

def replacement():
    original()
    print "notify somebody..."

My issue is I think I'm getting my scoping etc., wrong as the output of the following code is not as expected. Maybe if I could ask is there a better way to do this? I want to stick with modifying the original cb has as it third party code and less places I change the better.

#!/usr/bin/python 

def a():
    print "a"

def b():
    print "b"

def c():
    print "c"


orig_fxn_cb = dict()

" basic name to function callback hash "
orig_fxn_cb['a'] = a
orig_fxn_cb['b'] = b
orig_fxn_cb['c'] = c

" for each call back routine in the hash append a final action to it "
def appendFxn(fxn_cb):
    appended_fxn_cb_new = dict()
    for i in orig_fxn_cb.keys():
        cb = fxn_cb[i]
        def fxn_tail():
            cb()
            print cb.__name__, "tail" 

        appended_fxn_cb_new[i] = fxn_tail
        appended_fxn_cb_new[i]()
    return appended_fxn_cb_new

" make up a modified callback hash "
xxx = appendFxn(orig_fxn_cb)
print xxx

for i in xxx:
    print xxx[i]()

If I may, your variable names aren't the most descriptive. That definitely makes it harder to keep track of what's going on. — jathanism
– jathanism, Commented Nov 20, 2012 at 23:49

Eric · Accepted Answer · 2012-11-20 23:53:24Z

4

Are you after:

from functools import wraps

def notifier(f):
    @wraps(f)
    def wrapped(*args, **kwargs)
        res = f(*args, **kwargs)
        print "notify somebody..."
        return res
    return wrapped

@notifier
def original():
    print "original work"

# or: original = notifier(original)

original()

original work
notify somebody...

answered Nov 20, 2012 at 23:53

Eric

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6 Comments

jathanism Over a year ago

Decorators to the rescue! BTW, you don't have to use functools.wraps, but it makes sure to preserve the docstring and argument signature of the wrapped (decorated) function, so it's a good thing to keep in mind.

Eric Over a year ago

@jathanism: "it makes sure to preserve the argument signature" - pretty sure that's not the case, and you have to get the decorator module for that. (Which works by generating python code as a string).

abarnert Over a year ago

It's worth noting that you can also call decorators explicitly. For example, when creating your dict from names to functions, you can do, e.g., d['foo'] = notifier(foo).

abarnert Over a year ago

@jathanism: Yes, Eric is right. Try it out: def foo(): pass, then bar=notifier(foo), then help(foo) shows foo(), while help(bar) shows bar(*args, **kwargs).

jathanism Over a year ago

Ah, thanks for pointing that out. So just the docstring. Got it! :)

|

Blckknght · Accepted Answer · 2012-11-21 00:02:18Z

1

The issue you are running into has to do with the cb variable that you're using in your appendFxn function. Your inner function definitions access this value by name, and so they all end up having a reference to the same inner callback function (whichever one came up last in the iteration).

You can fix this with a default argument in your wrapper function:

def appendFxn(fxn_cb):
    appended_fxn_cb_new = dict()
    for i in orig_fxn_cb.keys():
        cb = fxn_cb[i]
        def fxn_tail(cb = cb):           # fix is here!
            cb()
            print cb.__name__, "tail" 

        appended_fxn_cb_new[i] = fxn_tail
        appended_fxn_cb_new[i]()
    return appended_fxn_cb_new

Using a default argument binds the current value of cb to the cb name within the inner function. That way, even when the original cb variable gets set to a new value, the existing function will keep the old one. (You could also make the function argument cb = fxn_cb[i] and get rid of the outer cb variable if you want.)

answered Nov 21, 2012 at 0:02

Blckknght

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1 Comment

David Kierans Over a year ago

Thanks. This fixed the issue I was having.

BoppreH · Accepted Answer · 2012-11-20 23:54:11Z

0

I don't know what's the expected output, it seems to work fine here. It replaces all functions in the dictionary with a copy that calls the original and then prints the original name and "tail". As it replaces the functions, it also executes them. In the end, you iterate over the resulting dictionary and execute them again.

Something that may give you unexpected behavior if you call appendFxn many times is the for i in orig_fxn_cb.keys(): line. It references the global orig_fxn_cb instead of the fxn_cb parameter passed, so further calls will not change the functions in the map.

answered Nov 20, 2012 at 23:54

BoppreH

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