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?

Answer 1

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) 

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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>>)