Calling member functions?

[Peter Wang]

"David Dawkins" <david_j_dawkins at spamless.hotmail.com> wrote:
> class  Handler:
>     def Callback(self):
>         print "Callback called"
> class Notifier:
>     def __init__(self, object, method):
>         self.m_object = object
>         self.m_method = method
>     def notify(self):
>         ## Now what??
>         self.m_method(self.m_object) ## naive attempt
> h = Handler()
> n = Notifier( h, h.Callback )
> h.notify()  # I want this to do h.Callback() in effect

(i'll assume you mean n.notify in the last line.)

when you instantiate n with "h.Callback", you're assigning a *bound*
function reference to n.m_method.  you can see this by just printing
out the value of n.m_method.  you'll get something like:

<method Handler.Callback of Handler instance at 0x80dba14>

thus, when you call m_method(), you don't have to provide an instance
for the "self" variable because the function is already bound to an
instance (specifically, to h).

if you change your code a little bit to the following:

class Notifier:
	def __init__(self, object, method):
		self.m_object = object
		self.m_method = method
	def notify(self):
		self.m_method(self.m_object) ## naive attempt
h = Handler()
n = Notifier(h, Handler.Callback)
n.notify()

this will also work.  notice that here, n is instantiated with an
*unbound* function, named Handler.Callback (which is different from
h.Callback).  in this case, if you want to call n.m_method(), you have
to provide an instance for the "self" parameter, so you have to pass
in self.m_object.

incidentally, if you print out the value of n.m_method in this
modified code, you get:

<unbound method Handler.Callback>

HTH,
peter