Implementing a callback in Python - passing a callable reference to the current function

Question

I want to implement the Observable pattern in Python for a couple of workers, and came across this helpful snippet:

class Event(object):
    pass

class Observable(object):
    def __init__(self):
        self.callbacks = []
    def subscribe(self, callback):
        self.callbacks.append(callback)
    def fire(self, **attrs):
        e = Event()
        e.source = self
        for k, v in attrs.iteritems():
            setattr(e, k, v)
        for fn in self.callbacks:
            fn(e)

Source: Here

As i understand it, in order to subscribe, I would need to pass a callback to the function that is going to be called on fire. If the calling function was a class method, presumably I could have used self, but in the absence of this - how could I directly get a callback that can be useful for the self.callbacks.append(callback) bit?

Solution

Any defined function can be passed by simply using its name, without adding the () on the end that you would use to invoke it:

def my_callback_func(event):
    # do stuff

o = Observable()
o.subscribe(my_callback_func)

---

Other example usages:

class CallbackHandler(object):
    @staticmethod
    def static_handler(event):
        # do stuff

    def instance_handler(self, event):
        # do stuff

o = Observable()

# static methods are referenced as <class>.<method>
o.subscribe(CallbackHandler.static_handler)

c = CallbackHandler()
# instance methods are <class instance>.<method>
o.subscribe(c.instance_handler)

# You can even pass lambda functions
o.subscribe(lambda event: <<something involving event>>)